NOJ [1122] Shameimaru's Candid Camera

• [1122] Shameimaru's Candid Camera

  • 问题描述
  • Shameimaru Aya(しゃめいまる·あや) likes to photograph others secretly. So she plans to set some photo point in the map.
    Once she has set all the photo point, she have to sign out it.
    Every coordinate of the map have to sign out the number of photo point nearby. The word "nearby" means the 9-cell matrix with the point in the center.
    You can see the image below:
    
    (The left top point is (0, 0)).
    
    Your task is to help Shameimaru to sign out every cell in the map. (Except the photo point)
    

    
    
  • 输入
  • This problem has several cases. The first line of each case contains two integers: N, M(0 < N, M <= 500), Means the width and the height of the map.
    Then follows a map larges (N, M). '-' indicates a space and '*' means a photo point.
    
  • 输出
  • For each case, output the map after signing. Leave a single line between each case.

  水题。。。。。。

  #include<stdio.h>
  #include<string.h>
  #define inf 0x3f3f3f
  char mat[510][510];
  int vis[510][510];
  
  int main()
  {
      int n,m,page=1; //n是宽度
      while(~scanf("%d%d",&n,&m))
      {
          if(page>1)
              printf("\n");
          page++;
          memset(vis,0,sizeof(vis));
          for (int i = 0; i < m; ++i)
              scanf("%s",&mat[i]);
          for(int i=0;i<m;i++)
              for(int j=0;j<n;j++)
                  if(mat[i][j]=='*')
                  {
                      vis[i][j]=inf;
                      if(i-1>=0 && mat[i-1][j]!='*')
                          vis[i-1][j]++;
                      if(i-1>=0 && j-1>=0 && mat[i-1][j-1]!='*')
                          vis[i-1][j-1]++;
                      if(i-1>=0 && j+1<n &&  mat[i-1][j+1]!='*')
                          vis[i-1][j+1]++;
                      if(j-1>=0 && mat[i][j-1]!='*')
                          vis[i][j-1]++;
                      if(j+1<n && mat[i][j+1]!='*')
                          vis[i][j+1]++;
                      if(i+1>=0 && mat[i+1][j]!='*')
                          vis[i+1][j]++;
                      if(i+1>=0 && j-1>=0 && mat[i+1][j-1]!='*')
                          vis[i+1][j-1]++;
                      if(i+1>=0 && j+1<n && mat[i+1][j+1]!='*')
                          vis[i+1][j+1]++;
                  }
          for(int i=0;i<m;i++)
          {
              for(int j=0;j<n;j++)
              {
                  if(vis[i][j]==inf)
                      printf("*");
                  else if(vis[i][j]==0)
                      printf("-");
                  else
                      printf("%d", vis[i][j]);
              }
              printf("\n");
          }
      }
      return 0;
  }