postype（error C2784:）

本文目录

• error C2784:
• 向心算法
• 马踏棋盘的核心代码（c语言）
• 用matlab如何讲图片的其中一部分进行裁剪，然后对裁剪的部分进行上下倒转之后再放回原图啊
• postype可以用推特登录吗
• 韩国软件postype怎么下载
• postype怎么下载
• MUGEN怎样给人物加上特效

error C2784:

楼主，错误如下：1.// e=(curstep,curpos,1); //改为下面3行的内容 e.ord=curstep; ////定义的postype e; 只是postype的对象，应该调用postype里面的成员进行赋值，上面错误的写法是函数赋值 e.seat=curpos; //// e.di=1; ////2.// if(curpos==end) return true; //改为下面1行的内容 if(curpos.x==end.x && curpos.y==end.y) return true; //原因同上，应该调用coordinate里面的成员才能进行比较还有二维数组的写法问题，具体的说在程序里，看看吧：#include 《iostream》#include 《stack》#include 《malloc.h》#include 《vector》#include 《map》#include 《algorithm》using namespace std;#define n 10int a=, //二维数组应该这样写，小数组里也该用括号 , , , , , , , , };int b;typedef struct{ int x; int y;}coordinate; typedef struct{ int ord; coordinate seat; int di;}postype;bool pass(coordinate epos){ if(b==1) return true; else return false;}void footprint(coordinate curpos){ b=0;}coordinate nextpos(coordinate curpos,int t){ switch (t) { case 1: curpos.x++; case 2: curpos.y++; case 3: curpos.x--; case 4: curpos.y--; } return curpos;}bool mazepath(coordinate start,coordinate end){ postype e; stack《postype》 pos; coordinate curpos; curpos=start; int curstep=1; do { if(pass(curpos)) { footprint(curpos);// e=(curstep,curpos,1); //改为下面3行的内容 e.ord=curstep; ////定义的postype e; 只是postype的对象，应该调用postype里面的成员进行赋值，上面错误的写法是函数赋值 e.seat=curpos; //// e.di=1; //// pos.push(e);// if(curpos==end) return true; //改为下面1行的内容 if(curpos.x==end.x && curpos.y==end.y) return true; //原因同上，应该调用coordinate里面的成员才能进行比较 curpos=nextpos(curpos,1); curstep++; } else { if(!pos.empty()) { pos.pop(); while (e.di==4&&!pos.empty()) { footprint(e.seat); pos.pop(); } if(e.di《4) { e.di++; pos.push(e); curpos=nextpos(e.seat,e.di); } } } }while(!pos.empty()); return false;}void main(){ coordinate start,end; int i,j; b; for(i=0;i《n;i++) { for(j=0;j《n;j++) cout《《a《《“ “; cout《《endl; } start.x=1; start.y=1; end.x=8; end.y=8; mazepath(start,end);}另外,团IDC网上有许多产品团购,便宜有口碑

向心算法

没听说过“迷宫向心算法“,你看看迷宫的全部代码吧!看对你有没有帮助.迷宫的代码如下:#include 《stdio.h》#include 《stdlib.h》 #include 《malloc.h》#include 《conio.h》#include 《dos.h》#define OK 1#define ERROR 0#define TRUE 1#define FALSE 0#define F9 0x43#define Esc 0x1b#define Del 0x53#define Home 0x47#define End 0x4f#define Space 0x20#define Up 0x48#define Down 0x50#define Left 0x4b#define Right 0x4d#define Enter 0x0d#define F2 0x3c#define F3 0x3d#define STACK_INIT_SIZE 200#define STACKINCREMENT 10typedef int Boolean;typedef int Status;typedef struct {int x;int y;} PosType;typedef struct {int ord;PosType seat;int di;} SElemType;typedef struct {int td;int foot;int mark;} MazeType;typedef struct {SElemType *base;SElemType *top;int stacksize;} Stack;int Maze;MazeType maze;PosType StartPlace;PosType EndPlace;int count;int m,n; 字串9 Boolean b_start=FALSE,b_end=FALSE;void CreatMaze(void);Status SaveMaze(char *filename);Status LoadMaze(char *filename);void Error(char *message);Status InitStack(Stack *s);Status DestroyStack(Stack *s);Status ClearStack(Stack *s);Boolean StackEmpty(Stack *s);int StackLength(Stack *s);Status Push(Stack *s,SElemType e);SElemType Pop(Stack *s,SElemType e);Status GetTop(Stack *s,SElemType *e);Status StackTraverse(Stack *s,Status (* visit)(SElemType *se));Boolean Pass(PosType curpos);void MarkPrint(PosType seat);void FootPrint(PosType curpos);PosType NextPos(PosType seat,int di);Status MazePath(PosType start,PosType end);void CreatMaze(void)/* Form the maze. */{void Error(char *message);Status SaveMaze(char *filename);Status LoadMaze(char *filename);int i,j;int x,y;char c;char savename;Boolean flag=FALSE,load=FALSE;clrscr(); 字串3printf(“Menu:\n\n“);printf(“1.Load Mazefile:(*.dd)\n\n“);printf(“2.Input Maze:\n\n“);printf(“Input your choice: “);do{c=getch();switch(c){case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’); break;case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’); break;case Esc: sleep(1); exit(1);default: break;}}while(c!=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’&&c!=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’) ;if(c==’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’){printf(“\n\nLoadName: “);scanf(“%s“,loadname);if(LoadMaze(loadname)){sleep(1); load=TRUE;}else { gotoxy(1,9); printf(“Load fail! “); } 字串1}if(!load){printf(“\nInput the maze’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’s size:\n“);printf(“\nInput Length :\n“);scanf(“%d“,&m);printf(“\nInput Width :\n“);scanf(“%d“,&n);if(m《4||n《4) Error(“Input“);if(m》30||n》20) Error(“Maze too large“);for(i=0;i《30;i++)for(j=0;j《20;j++)Maze=2;StartPlace.x=0;StartPlace.y=0;EndPlace.x=0;EndPlace.y=0;clrscr();printf(“\n“);for(i=1;i《=n;i++){for(j=1;j《=m;j++){printf(“ #“);Maze=0;}printf(“\n“);}}gotoxy(65,5);printf(“’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’#’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’:Wall“);gotoxy(65,7);printf(“Start:Home“);gotoxy(65,9);printf(“End:End“);gotoxy(65,11);printf(“Delete Wall:Del“); 字串3gotoxy(65,13);printf(“Enter Wall:Enter“);gotoxy(65,15);printf(“Save Maze:F2“);gotoxy(65,17);printf(“Complete:F9“);gotoxy(65,19);printf(“Exit:Esc“);gotoxy(4,3);x=4;y=3;do{c=getch();switch(c){case Up: if(y》3) { y--; gotoxy(x,y); }break;case Down: if(y《n) { y++; gotoxy(x,y); }break;case Left: if(x》4) { x-=2; gotoxy(x,y); }break;case Right: if(x《2*m-2) { x+=2; gotoxy(x,y); }break;case Del: if(y-2==StartPlace.y&&x/2-1==StartPlace.x) b_start=FALSE;if(y-2==EndPlace.y&&x/2-1==EndPlace.x) b_end=FALSE;putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’ ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’); Maze=1; gotoxy(x,y);break;case Enter: if(y-2==StartPlace.y&&x/2-1==StartPlace.x) break;if(y-2==EndPlace.y&&x/2-1==EndPlace.x) break;putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’#’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’); Maze=0; gotoxy(x,y); 字串5break;case Home: if(Maze&&!b_start){StartPlace.x=x/2-1;StartPlace.y=y-2;putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’S’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);gotoxy(x,y);b_start=TRUE;}break;case End: if(Maze&&!b_end){EndPlace.x=x/2-1;EndPlace.y=y-2;putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’E’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);gotoxy(x,y);b_end=TRUE;}break;case Esc: gotoxy(2,22); printf(“exit“); sleep(1); exit(1);case F9: if(b_start&&b_end) flag=TRUE; break;case F2: gotoxy(2,22);printf(“Savename:“);scanf(“%s“,savename);gotoxy(2,22);if(SaveMaze(savename)) printf(“Save OK! “);else printf(“Save fail! “);sleep(1);gotoxy(2,22);printf(“ “);gotoxy(x,y);break;default: break;}}while(!flag);for(i=0;i《30;i++)for(j=0;j《20;j++) 字串2 {maze;maze.mark=0;maze.foot=0;}}Status LoadMaze(char *file)/* The maze has been loaded. */{FILE *fp;char *buffer;char ch;int i=0,j,k;Boolean len=FALSE,wid=FALSE;if((fp=fopen(file,“r“))==NULL)return ERROR;buffer=(char *)malloc(600*sizeof(char));ch=fgetc(fp);while(ch!=EOF){buffer=ch;i++;ch=fgetc(fp);}m=30;n=20;for(i=0;i《600;i++){j=i/30; k=i%30;if(buffer==’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’&&!len){ m=i; len=TRUE; }if(k==0&&buffer==’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’&&!wid){ n=j; wid=TRUE; }switch(buffer){case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’0’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: Maze=0; break;case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: Maze=1; break; 字串4 case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: Maze=2; break;case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’3’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: Maze=1;StartPlace.x=k;StartPlace.y=j;b_start=TRUE;break;case ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’4’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’: Maze=1;EndPlace.x=k;EndPlace.y=j;b_end=TRUE;break;default : break;}}fclose(fp);clrscr();for(i=0;i《30;i++)for(j=0;j《20;j++){maze;maze.foot=0;maze.mark=0;if(Maze==0){gotoxy(2*i+2,j+2);putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’#’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);}}gotoxy(2*StartPlace.x+2,StartPlace.y+2);putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’S’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);gotoxy(2*EndPlace.x+2,EndPlace.y+2);putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’E’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);return OK; 字串2}Status SaveMaze(char *filename)/* The maze has been saved. */{FILE *fp;char *buffer;int i,j,k;fp=fopen(filename,“wb“);buffer=(char *)malloc(600*sizeof(char));for(i=0;i《600;i++){j=i/30; k=i%30;switch(Maze){case 0: buffer=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’0’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’; break;case 1: buffer=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’; break;case 2: buffer=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’2’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’; break;default : Error(“Write“); break;}if(k==StartPlace.x&&j==StartPlace.y) buffer=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’3’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’;if(k==EndPlace.x&&j==EndPlace.y) buffer=’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’4’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’;}fwrite(buffer,600,1,fp);free(buffer);fclose(fp);return OK;}void Error(char *message) 字串6{clrscr();fprintf(stderr,“Error:%s\n“,message);exit(1);} /* Error */Status InitStack(Stack *s)/* The stack s has been created and is initialized to be empty. */{s-》base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));if(!s-》base) Error(“Overflow“);s-》top=s-》base;s-》stacksize=STACK_INIT_SIZE;return OK;} /* InitStack */Status DestroyStack(Stack *s)/* The stack s has been destroyed. */{s-》top=NULL;s-》stacksize=0;free(s-》base);s-》base=NULL;return OK;} /* DestroyStack */Status ClearStack(Stack *s)/* The stack has been clear to be maximum. */{s-》top=s-》base;s-》stacksize=STACK_INIT_SIZE;return OK;} /* ClearStack */Boolean StackEmpty(Stack *s)/* Check if the stack s is empty. */{if(s-》top==s-》base) return TRUE;else return FALSE;} /* StackEmpty */ 字串8 int StackLength(Stack *s)/* Gain the length of the stack s. */{if(s-》top》s-》base) return (int)(s-》top-s-》base);else return 0;} /* StackLength */Status Push(Stack *s,SElemType e)/* The element e has been pushed into the stack s. */{if(s-》top-s-》base》=s-》stacksize){s-》base=(SElemType *)realloc(s-》base,(s-》stacksize+STACKINCREMENT)*sizeof(SElemType));if(!s-》base) Error(“Overflow“);s-》top=s-》base+s-》stacksize;s-》stacksize+=STACKINCREMENT;}*s-》top++=e;return OK;} /* Push */SElemType Pop(Stack *s,SElemType e)/* The element e has been removed from the stack s. */{if(s-》top==s-》base) Error(“Pop“);e=*--s-》top;return e;} /* Pop */Status GetTop(Stack *s,SElemType *e)/* The element e has got to the top of the stack s.*/{if(s-》top==s-》base) Error(“GetTop“); 字串4 *e=*(s-》top-1);return OK;} /* GetTop *//* Traverse the stack s using ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’visiting’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’ function. *//* Status StackTraverse(Stack *s,Status (* visit)(SElemType *se)){SElemType p;int result;if(s-》top==s-》base) return ERROR;p=s-》base;while(!(p==s-》top)){result=(*visit)(p);p++;}return OK;} */Boolean Pass(PosType curpos)/* Check if the current position can be passed. */{if(maze.td==1&&maze.mark==0)return TRUE;else return FALSE;} /* Pass */void MarkPrint(PosType seat)/* Mark the position seat. */{maze.mark=-1;/* Marking ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’-1’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’ symbolize the current position cannot be put. */} /* MarkPrint */void FootPrint(PosType curpos) 字串5 /* The foot of the curpos of the maze has been set ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’true’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’. */{maze.foot=1;} /* FootPrint */PosType NextPos(PosType seat,int di){switch(di){case 1: seat.y++; return seat; /* Eastward */case 2: seat.x++; return seat; /* Southward */case 3: seat.y--; return seat; /* Westward */case 4: seat.x--; return seat; /* Northward */default: seat.x=0; seat.y=0; return seat;}} /* NextPos *//* The key to the program. *//* Pre: The maze array & the startplace & the endplace.Post: Find the one traverse of the maze and perform the mazepath.Uses: The ADT stack class.*/Status MazePath(PosType start,PosType end){PosType curpos;int curstep;SElemType e;Stack *s,stack;stack.base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));if(!stack.base) Error(“Overflow“); 字串8 stack.top=stack.base;stack.stacksize=STACK_INIT_SIZE;s=&stackcurpos=start;curstep=1;do{if(Pass(curpos)){FootPrint(curpos);e.ord=curstep; e.seat=curpos; e.di=1;gotoxy((curpos.y+1)*2,curpos.x+2);putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’@’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);delay(8000); /* pospone time. */Push(s,e);if(curpos.x==end.x&&curpos.y==end.y) /* Proceed recursively. */{DestroyStack(s);return TRUE;}curpos=NextPos(curpos,1); /* Try next position. */curstep++;}else{if(!StackEmpty(s)){e=Pop(s,e); /* Removed e from s. */while(e.di==4&&!StackEmpty(s)) /* Four directions have been checkedand s is not empty. */{MarkPrint(e.seat);gotoxy((e.seat.y+1)*2,e.seat.x+2);putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’@’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);delay(8000); /* Pospone time. */gotoxy((e.seat.y+1)*2,e.seat.x+2); 字串1 putch(’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’ ’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’);e=Pop(s,e); /* Remove e from s. */curstep--;}if(e.di《4) /* The current position hasnot been checked. */{e.di++;Push(s,e); /* Insert e into s. */curpos=NextPos(e.seat,e.di); /* Try next position. */}}}}while(!StackEmpty(s));DestroyStack(s);return FALSE;} /* MazePath */void main(){PosType start,end;CreatMaze();start.x=StartPlace.y;start.y=StartPlace.x;end.x=EndPlace.y;end.y=EndPlace.x;if(MazePath(start,end)){gotoxy(2,22);printf(“Path found\n“);}else{gotoxy(2,22);printf(“Path not found\n“);}getch();clrscr();}

马踏棋盘的核心代码（c语言）

#define STACK_INIT_SIZE 100#define STACKINCREMENT 10#define OVERFLOW -2#define OK 1#include《malloc.h》#include《stdio.h》#include《stdlib.h》int Board={0};int HTry1={2,-1,1,-2,2,1,-1,-2};int HTry2={1,2,2,1,-1,-2,-2,-1};typedef struct{int i;int j;}PosType;typedef struct{int ord;PosType seat;int di;}SElemType;typedef struct{SElemType *base;SElemType *top;int stacksize;}SqStack;int InitStack(SqStack *s1){(*s1).base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));if(!(*s1).base) exit(OVERFLOW);(*s1).top=(*s1).base;(*s1).stacksize=STACK_INIT_SIZE;return(OK);}SElemType Pop(SqStack *s,SElemType e){e=*--(*s).top;return e;}int Push(SqStack *s1,SElemType e){if((*s1).top-(*s1).base》=(*s1).stacksize){(*s1).base=(SElemType *)realloc((*s1).base,((*s1).stacksize+STACKINCREMENT)*sizeof(SElemType));if(!(*s1).base) exit(OVERFLOW);(*s1).top=(*s1).base+(*s1).stacksize;(*s1).stacksize+=STACKINCREMENT;}*(*s1).top++=e;return OK;}int StackEmpty(SqStack *s){if((*s).base==(*s).top)return(1);elsereturn(0);}int curstep=0;int Pass(PosType s){if((Board==0)&&(s.i《=7)&&(s.i》=0)&&(s.j《=7)&&(s.j》=0))return(1);elsereturn(0);}PosType NextPos(PosType s,int i){s.i=s.i+HTry1;s.j=s.j+HTry2;return(s);}void horsesteps(int Board,PosType start){int k,j;SqStack S;SElemType e;PosType curpos=start;InitStack(&S);do{if(Pass(curpos)){curstep++;Board=curstep;e.seat=curpos;e.ord=curstep;e.di=1;Push(&S,e);if(curstep==64)break;elsecurpos=NextPos(curpos,1);}//ifelse{if(!StackEmpty(&S)){Pop(&S,e);if(e.di==8) Board=0;while(e.di==8&&!StackEmpty(&S)){e=Pop(&S,e);if(e.di==8) Board=0;curstep=e.ord;}//whileif(e.di《8){e.di++;Push(&S,e);curpos=NextPos(e.seat,e.di);}//if}//if}//else}while(!StackEmpty(&S));if(StackEmpty(&S)){printf(马儿从这个初始位置不能踏遍棋盘\n);printf(请按任意键推出本程序\n);getchar();exit(OVERFLOW);}for(j=0;j《8;j++){printf(\n);for(k=0;k《8;k++)printf(%3d,Board);}// for}//函数结束void main(){int k,j;PosType t;char a,b;printf(请输入马儿的初始位置\n);fflush(stdin);scanf(%c%d,%d%c,&a,&k,&j,&b);t.i=k;t.j=j;printf(马儿走的路线为\n);horsesteps(Board,t);}小弟只能写出自己的回溯法的算法，效率很低，有时要21分钟出结果 ，但思路是对的。【原创】我给出我写的试探法代码#include 《iostream》using namespace std;typedef struct _point{int x;int y;}point;class Horse{public:Horse(int n);~Horse();void MoveNext(int hang,int lie);void Print();public:int shu;int qipan;point pt;int ci;};Horse::Horse(int n){ci=0;shu=n;for(int i=1;i《=shu;i++)for(int j=1;j《=shu;j++)qipan=0;}Horse::~Horse(){}void Horse::Print(){if(pt.x==shu*shu){ci++;cout《《ci《《 Yes!《《endl;for(int i=1;i《=shu;i++)for(int j=1;j《=shu;j++){cout《《qipan《《’\t’;if(j==shu)cout《《endl;}}}void Horse::MoveNext(int hang,int lie){if(hang==1&&lie==1&&qipan==0){pt.x=1;pt.x=hang;pt.y=lie;qipan.x;}if(hang-1》0&&lie-2》0&&qipan==0){pt.x++;hang=hang-1;lie=lie-2;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie); //返回点}hang=pt.x;lie=pt.y;if(hang+1《=shu&&lie-2》0&&qipan==0){pt.x++;hang=hang+1;lie=lie-2;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang-2》0&&lie-1》0&&qipan==0){pt.x++;hang=hang-2;lie=lie-1;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang+2《=shu&&lie-1》0&&qipan==0){pt.x++;hang=hang+2;lie=lie-1;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang-2》0&&lie+1《=shu&&qipan==0){pt.x++;hang=hang-2;lie=lie+1;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang+2《=shu&&lie+1《=shu&&qipan==0){pt.x++;hang=hang+2;lie=lie+1;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang-1》0&&lie+2《=shu&&qipan==0){pt.x++;hang=hang-1;lie=lie+2;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;if(hang+1《=shu&&lie+2《=shu&&qipan==0){pt.x++;hang=hang+1;lie=lie+2;pt.x=hang;pt.y=lie;qipan.x;MoveNext(hang,lie);//返回点}hang=pt.x;lie=pt.y;Print();{qipan=0;pt.x--;}}int main(){printf(Input the num:);int n;scanf(%d,&n);Horse h(n);h.MoveNext(1,1);return 0;}翘了一学期的数据结构，赶工补作业中，我用的是贪心算法，速度不错，秒出。。#include《iostream》#define M 12#define D 8using namespace std;int pathnum(int x,int y);void initdata();int pathnum(int x,int y);void findway(int x , int y );int board={0,0};int HTry1={ -2, -1, 1, 2, 2, 1, -1, -2 };int HTry2={ 1, 2, 2, 1, -1, -2, -2, -1 };void initdata(){int i , j;for ( i=0 ; i《2 ; i++ ){for ( j=0;j《M;j++){board=1;board=1;board=1;board=1;}}}int main(){void findway(int x , int y );int startx,starty,i,j;cout《《请输入初始坐标值《《endl;initdata();cin》》startx》》starty;findway(startx,starty);for(i=2;i《M-2;i++){//以图形方式显示行走步数for(j=2;j《M-2;j++)printf(%4d,board);cout《《endl《《endl;}return 0;}void findway(int x , int y ){x=x+1;y=y+1;int stepnum=1;int num;int i,min,minx,miny;board=stepnum;for ( stepnum=2; stepnum《=(M-4)*(M-4) ;stepnum++ ){min=8;for (i=0;i《D;i++){if (board ==0){num );if (num《min){min=num;minx = x+HTry1;miny = y+HTry2;}}}x=minx;y=miny;board=stepnum;}}int pathnum(int x,int y){//返回当前位置可走的方向数目int count=0 ;for (int i=0;i《D;i++){if (board==0){count++;}}return count ;}

用matlab如何讲图片的其中一部分进行裁剪，然后对裁剪的部分进行上下倒转之后再放回原图啊

使用 imcrop 时，返回两个值即可简单实现：

= imcrop(im)

上面的变量意义：im 是需要处理的原始图片，jm 是图片裁剪部分，rect 是坐标

其中，rect 是一个包含 4 个元素的数组，前两个元素是裁剪部分左上角的坐标 x 和 y，后两个元素是裁剪部分的大小，格式为 。注意，x 方向是宽，y方向是高，而 Matlab 是列优先索引，因此实际索引时需要将 y 方向作为第一维；

你的代码可以这样写:

%read image

img = imread(’baby.jpg’); 

figure;

subplot(1,2,1)

%show image

imshow(img); 

title(’ORIGINAL IMAGE’);

%crop image 

= imcrop(img);

rect = num2cell(round(rect));

= deal(rect{:}); 

%Upside down 

imud = im;

imud(ypos:ypos+ylen, xpos:xpos+xlen, :) = flipud(imud(ypos:ypos+ylen, xpos:xpos+xlen, :));

subplot(1,2,2);

imshow(imud);

代码示例

相比原代码，变动主要在 crop image 部分

运行结果

另附一个具有参数检查功能的函数如下，可实现根据坐标进行部分翻转，首个参数为图片变量，第二个参数（可选）决定坐标的类型，’a’ 是绝对坐标，表示输入数组下标，’r’ 是相对坐标，表示坐标在全图的比例（范围是0-1），具体代码如下：

function ret = partflipud(varargin)

%PARTFLIPUD Flip part of an array up to down.

%% Check Number of Arguments

narginchk(1, 6);

nargoutchk(0, 1);

%% Check Validation of Arguments im

im = varargin{1};

validateattributes(im, {’numeric’}, {’3d’, ’nonnan’});

args = nargin - 1;

if args == 0

    ret = flipud(im);

    return

end

%% Processing Other Arguments

if args 》 0

    abspos = false; %’absolute’

    if ischar(varargin{2}) || isstring(varargin{2})

        postype = varargin{2};

        if any(strcmpi(postype, {’absolute’, ’a’, ’abs’}))

            abspos = true;

        elseif any(strcmpi(postype, {’relative’, ’r’, ’rel’}))

            abspos = false;

        else

            error();

        end

        args = args - 1;

        if args == 0

            error(’Missing argument position.’);

        end

    end

    if args == 1

        pos = varargin{3};

        args = numel(pos);

    else

        pos = varargin(end-args+1:end);

    end

    if args ~= 4

        error();

    end

    = pos{:}; 

    if x1 》 x2

        error(’coordinate x1 greater than x2’);

    elseif y1 》 y2

        error(’coordinate y1 greater than y2’);

    end

    if ~abspos

        = size(im);

        x1 = floor(x1 * xlen);

        y1 = floor(y1 * ylen);

        x2 = ceil (x2 * xlen);

        y2 = ceil (y2 * ylen);

    end

end

%% Flip Specified Part of Array

ret = im;

ret(y1:y2, x1:x2, :) = flipud(ret(y1:y2, x1:x2, :));

end % function partflipud

参数检查部分

函数第 4-14 行检查输入和输出参数的合法性，支持输入灰度图像和 RGB/RGBA 格式图像；

坐标类型判断

函数第 15-31 行判断输入坐标的类型，绝对坐标可以用 ’absolute’, ’a’, ’abs’ 表示，绝对坐标可以用 ’relative’, ’r’, ’rel’，默认使用相对坐标；

坐标参数预处理和图片裁剪翻转处理

函数余下部分进行坐标处理和图像裁剪翻转；

postype可以用推特登录吗

可以。postype可以用推特登录。postype是一个韩国同人文网站。Twitter（推特）是一家美国社交网络及微博客服务的公司，致力于服务公众对话。Twitter可以让用户更新不超过140个字符的消息（除中文、日文和韩语外已提高上限至280个字符），这些消息也被称作推文(Tweet)，Twitter被形容为互联网的短信服务。这个服务是由杰克·多西(JackDorsey)在2006年3月与合伙人共同创办并在当年7月启动的。

韩国软件postype怎么下载

下载步骤如下：1、首先打开手机，右滑进入主界面。2、其次选择该界面中的应用商店，打开后在搜索栏内搜索“韩国软件postype”。3、最后在该软件显示出来时点击下载即可。

postype怎么下载

postype下载办法如下：1、在网站搜索postype会显示官方网站，点击官方网站。2、进入网站最醒目的地方，出现下载postype安卓版按钮点击它。3、百度指引下载提示，在这个页面有一个下载到电脑。4、然后，显示下载框，选择自己要下载的文件路径，完成后点击下载按钮。5、等待下载完成，打开对应的文件夹，打开QQ后，在手机和电脑同时登录，选择发送到我的安卓手机。当然也有很多其他的办法传输。6、发送完成，打开手机QQ，显示刚刚发送过来的应用安装包文件，点击它。7、然后就是安装步骤，直接安装即可，这样，安卓手机就可以使用postype了。

MUGEN怎样给人物加上特效

1、用记事本打开人物的CNS文件 查找 Statedef -2 （每个CNS文本都找找，如果确定找不到就自己手动添加一个 ）。2、自己整理好素材，载入SFF图包并且好相对应的Air动作。3、在-2下面加入光圈代码（假设你的光圈动画的编号是9997）。type = Explodtriggerall = !ishelpertriggerall = anim = 0triggerall = time 》= 15trigger1 = NumExplod(9997) = 0 ；括号里填写光圈动画代码anim = 9997 ；对应填写光圈代码id = 9997 ；填写光圈图片组代码sprpriority = -5postype = p1 ；跟随P1移动bindtime = -1pos = 0,-40 ；出现时的坐标 X轴越大越靠右，Y轴越小越靠上 以此类推under = 0ownpal = 1persistent = 0removetime = -1ignorehitpause = 99999999scale = 0.8,0.8SuperMoveTime = 9999PauseMoveTime = 9999