HDU6113度度熊的01世界（深搜）

度度熊的01世界

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1389    Accepted Submission(s): 526

Problem Description

度度熊是一个喜欢计算机的孩子，在计算机的世界中，所有事物实际上都只由0和1组成。

现在给你一个n*m的图像，你需要分辨他究竟是0，还是1，或者两者均不是。

图像0的定义：存在1字符且1字符只能是由一个连通块组成，存在且仅存在一个由0字符组成的连通块完全被1所包围。

图像1的定义：存在1字符且1字符只能是由一个连通块组成，不存在任何0字符组成的连通块被1所完全包围。

连通的含义是，只要连续两个方块有公共边，就看做是连通。

完全包围的意思是，该连通块不与边界相接触。

 

Input

本题包含若干组测试数据。
每组测试数据包含：
第一行两个整数n,m表示图像的长与宽。
接下来n行m列将会是只有01组成的字符画。

满足1<=n,m<=100

 

Output

如果这个图是1的话，输出1；如果是0的话，输出0，都不是输出-1。

 

Sample Input

32 32000000000000000000000000000000000000000000011111111000000000000000000000001111111111100000000000000000000011111111111100000000000000000001111111111111100000000000000000011111100011111000000000000000001111100000011110000000000000000011111000000111110000000000000000111110000000111110000000000000011111100000001111100000000000000111111000000001111100000000000001111110000000001111000000000000011111100000000011111000000000000111110000000000111100000000000001111000000000001111000000000000011110000000000011110000000000000111100000000000011100000000000000111100000000000111000000000000001111000000000001110000000000000011110000000000011100000000000001111000000000011110000000000000011110000000000111100000000000000011100000000001111000000000000000111110000011111110000000000000001111100011111111000000000000000011111111111111100000000000000000011111111111111000000000000000001111111111111000000000000000000001111111111100000000000000000000001111111000000000000000000000000011111000000000000000000000000000000000000000000000000032 3200000000000000000000000000000000000000000000000011111100000000000000000000000000111111100000000000000000000000011111111000000000000000000000001111111110000000000000000000000001111111100000000000000000000000011111111000000000000000000000001111111100000000000000000000000011111110000000000000000000000001111111100000000000000000000000011111111100000000000000000000000111111111000000000000000000000001111111100000000000000000000000111111100000000000000000000001111111111000000000000000000001111111111111000000000000000000111111111111110000000000000000001111111111111100000000000000000011111111111110000000000000000000000011111111110000000000000000000000000011111100000000000000000000000001111111000000000000000000000001111111100000000000000000000000001111111100000000000000000000000011111111000000000000000000000000111111111000000000000000000000001111111110000000000000000000000000111111110000000000000000000000000011111111110000000000000000000000111111111100000000000000000000000111111111000000000000000000000000000000000000003 3101101011

 

Sample Output

01-1

 

Source

2017"百度之星"程序设计大赛 - 初赛（A）

 

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解题思路：

上下左右搜索0的块和1的块，当0的块数为1，1的块数为1时，数字是1；

当1的块数是1，0的块数为0时，数字是0；

其他情况都是-1。

#include<stdio.h>  #include<stdlib.h>  #include<string.h>  using namespace std;  int n,m;  char map[120][120];  int vis[120][120];  int dir[4][2]={1,0,-1,0,0,1,0,-1};  int jude(int x,int y)  {      if(x<0||x>n||y<0||y>m) return 0;      return 1;    }  void dfs(int x,int y,char ch)  {      vis[x][y]=1;      for(int i=0;i<4;i++)      {          int xx=x+dir[i][0];          int yy=y+dir[i][1];          if(!jude(xx,yy)) continue;          if(vis[xx][yy]) continue;          if(map[xx][yy]!=ch) continue;          dfs(xx,yy,ch);      }  }  int main()  {      while(scanf("%d%d",&n,&m)!=EOF)      {          for(int i=0;i<=105;i++)          {             for(int j=0;j<=105;j++)             {                 map[i][j]='0';             }          }          for(int i=1;i<=n;i++)          {              scanf("%s",map[i]+1);              map[i][1+m]='0';          }          n++,m++;          memset(vis,0,sizeof(vis));          int count1=0,count2=0;          for(int i=0;i<=n;i++)          {              for(int j=0;j<=m;j++)              {  if(vis[i][j]) continue;                  if(map[i][j]=='0') count1++;                  else count2++;                  dfs(i,j,map[i][j]);                  /*for(int ii=1;ii<=n;ii++) {                for(int jj=1;jj<=m;jj++){                printf("%d",vis[ii][jj]);                }                printf("\n");                }*/            }          }          if(count1==1&&count2==1) puts("1");          else if(count1==2&&count2==1) puts("0");          else puts("-1");      }  }