Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21594 | Accepted: 10873 |
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure
out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
题目大致思路就是替换W周围所有W。。
从任意的W开始，不停地把邻接的部分用'.'代替。1次DFS后与初始的这个W连接的所有W就都被替换成了'.'，因此直到图中不再存在W为止，总共进行DFS的次数就是答案了。8个方向共对应了8种状态转移，每个格子作为DFS的参数至多被调用一次，所以复杂度为O(8×N×M)=O(N×M)。
#include<stdio.h>
char f[100][101];
int N,M;
void dfs(int x,int y)
{
f[x][y]='.';
int dx,dy;
for(dx=-1;dx<=1;dx++){
for(dy=-1;dy<=1;dy++){
int nx=x+dx,ny=y+dy;
if(nx>=0&&nx<N&&ny>=0&&ny<M&&f[nx][ny]=='W')dfs(nx,ny);
}
}
return;
}
int main()
{
void dfs(int x,int y);
// freopen("C:\\Users\\兆豫\\Desktop\\acm.txt","r",stdin);
while(scanf("%d%d",&N,&M)!=EOF){
int i,j,count=0;
for(i=0;i<N;i++)
// for(j=0;j<M;j++)
scanf("%s",f[i]);//无语，开始一直找不到错误，原来题目要求输入每行之间没有空格。
for(i=0;i<N;i++){
for(j=0;j<M;j++){
if(f[i][j]=='W'){
dfs(i,j);
count++;
}
}
}
printf("%d\n",count);
}
return 0;
}