poj 1544 A Puzzling Problem

poj 1544 A Puzzling Problem

#include <stdio.h>  #include <string.h>  const int N = 20;  char g[N][5][5];  int n, m[N][2], flag, vis[5][5], t, sum;  void init() {      flag = sum = 0;      memset(vis, 0, sizeof(vis));      t++;  }  void dfs(int cur) { //cur表示当前    if (cur == n + 1) { //如果但前 == n+1则返回结果为真        flag = 1;          for (int i = 0; i < 4; i++) { //打印结果 for (int j = 0; j < 4; j++)       printf("%d", vis[i][j]);   printf("n");          }          return;      }      if (flag)  //flag == true剪枝        return;      for (int i = 0; i <= 4 - m[cur][0]; i++) { //行        for (int j = 0; j <= 4 - m[cur][1]; j++) {//列   int flag1 = 1;//标记没有被覆盖 for (int p = 0; p < m[cur][0]; p++) { //行      for (int q = 0; q < m[cur][1]; q++) //列         if (flag1 && !vis[i + p][j + q] && g[cur][p][q] == '1')    vis[i + p][j + q] = cur;         else if (flag1 && g[cur][p][q] == '0')    continue;   else    flag1 = 0;   }   if (flag1)       dfs(cur + 1);   for (int p = 0; p < 4; p++)       for (int q = 0; q < 4; q++)if (vis[p][q] == cur)    vis[p][q] = 0;          }      }  }  int main() {      while (scanf("%d", &n) && n) {  //输入n个方块        init();          for (int i = 1; i <= n; i++) { scanf("%d%d", &m[i][0], &m[i][1]); //输入行和列  getchar();   for (int j = 0; j < m[i][0]; j++) { //统计     for (int k = 0; k < m[i][1]; k++) {scanf("%c", &g[i][j][k]); //第i个的图          sum += g[i][j][k] - '0';       }       getchar();   }          }          if (t != 1)  printf("n");          if (sum == 16)  //如果sum == 16 dfs dfs(1);          if (!flag)   printf("No solution possiblen");      }      return 0;  }