#include <stdlib.h>#include <time.h>//圖片類class picture{public: int type;//圖片的編號(hào)，共有n種，從0到n-1 bool visible;//圖片是否可見(jiàn) int x;//圖片位置的橫坐標(biāo) int y;//圖片位置的綜坐標(biāo)};//整個(gè)圖由8行10列組成，每個(gè)單元格是一張小圖片const int pNum = 10;const int pType = 8;static picture p[pType][pNum];//進(jìn)入新一關(guān)void newStage(){ srand(time(0)); int i,j; for(i = 0;i < pType;++i) for(j = 0;j < pNum;j++)  p[i][j].visible = false; int x,y; for (i = 0;i < pType - 1;++i) for(j = 0;j < pNum;++j) {  bool re = true;  while (re)  {  x = rand() % pType;  y = rand() % pNum;  if (p[x][y].visible == false)  {   p[x][y].type = i;   p[x][y].visible = true;   p[x][y].x = x;   p[x][y].y = y;   re = false;  }  } } //處理剩余的最后一種圖片 for (i = 0;i < pType;++i)  for(j = 0;j < pNum;++j)  {  if (p[i][j].visible == false)  {   p[i][j].type = pType - 1;   p[i][j].visible = true;   p[i][j].x = i;   p[i][j].y = j;  }  }} //在a、b兩點(diǎn)之間畫(huà)線void drawLine(picture a,picture b){ }//判斷圖片a和b之間能否通過(guò)一條直線相連（a和b之間有0個(gè)轉(zhuǎn)角）bool matchDirect(picture a,picture b){ if(!(a.x == b.x || a.y == b.y)) return false; //a、b的橫坐標(biāo)相同時(shí) bool yMatch = true; if(a.x == b.x) { if(a.y > b.y) {  for(int i = b.y + 1;i < a.y;++i)  {  if(p[a.x][i].visible == true)   yMatch = false;  } } if(b.y > a.y) {  for(int i = a.y + 1;i < b.y;++i)  {  if(p[a.x][i].visible == true)   yMatch = false;  } } } //a、b的縱坐標(biāo)相同時(shí) bool xMatch = true; if(a.y == b.y) { if(a.x > b.x) {  for(int i = b.x + 1;i < a.x;++i)  {  if(p[i][a.y].visible == true)   xMatch = false;  } } if(b.x > a.x) {  for(int i = a.x + 1;i < b.x;++i)  {  if(p[i][a.y].visible == true)   xMatch = false;  } } } return (xMatch && yMatch);}//判斷圖片a和b之間是否可以通過(guò)一個(gè)轉(zhuǎn)角的折線相連bool matchOneCorner(picture a,picture b){ if (p[a.x][b.y].visible == false && matchDirect(a,p[a.x][b.y]) && matchDirect(p[a.x][b.y],b)) { drawLine(a,p[a.x][b.y]); drawLine(p[a.x][b.y],b); return true; } if (p[b.x][a.y].visible == false && matchDirect(a,p[b.x][a.y]) && matchDirect(p[b.x][a.y],b)) { drawLine(a,p[b.x][a.y]); drawLine(p[b.x][a.y],b); return true; } return false;}//判斷圖片a和b之間是否可以通過(guò)兩個(gè)轉(zhuǎn)角的折線相連bool matchTwoCorner(picture a,picture b){ int i,j; for(i = a.x - 1,j = a.y;i >= 0;--i) { if(p[i][j].visible == true)  break; else if(matchOneCorner(b,p[i][j])) {  drawLine(a,p[i][j]);  return true; } } for (i = a.x + 1,j = a.y;i < pNum;++i) { if(p[i][j].visible == true)  break; else if(matchOneCorner(b,p[i][j])) {  drawLine(a,p[i][j]);  return true; } } for(i = a.x,j = a.y - 1;j >= 0;--j) { if(p[i][j].visible == true)  break; else if(matchOneCorner(b,p[i][j])) {  drawLine(a,p[i][j]);  return true; } } for(i = b.x,j = b.y + 1;j < pType;++j) { if(p[i][j].visible == true)  break; else if(matchOneCorner(b,p[i][j])) {  drawLine(a,p[i][j]);  return true; } } return false;}//判斷a和b能否相連，條件是a和b的類型相同，且a和b之間的連線拐角數(shù)<=2個(gè)bool match(picture a,picture b){ if(a.type != b.type) return false; if(matchDirect(a,b)) { drawLine(a,b); return true; } else if(matchOneCorner(a,b)) return true; else if(matchTwoCorner(a,b)) return true; return false;}