POJ - 2386 （dfs简单应用）

POJ - 2386 （dfs深度优先搜索）

题目正文：
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.
//由于最近的降雨，许多地方的水都汇集在农民约翰的田地里，它的代表是一个矩形的NxM(1<=N<=100；1<=M<=100)平方。每个广场都有水(‘W’)或旱地(‘.’)。农夫约翰想知道他的田地里形成了多少池塘。池塘是一组连在一起的正方形，里面有水，其中一个正方形被认为与它的八连通 //

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

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.
//有三个池塘：一个在左上角，一个在左下角，一个在右边。

题目解析：

对于每一个W，它周围的W(浅绿色)都算是跟它连在一起的，直到不能再继续连下去，则算一个池塘。
因此

sample output 的3就是这么来的。
很明显就是简单的dfs,这种地图的最好做了。

AC代码如下：

#include<stdio.h>
char field[100][100];   //地图
int N,M;
void dfs(int i,int j)
{   int x,y;int nx,ny;field[i][j]='.';       //表示搜索过，进行替换for(x=-1;x<=1;x++)for(y=-1;y<=1;y++)   {nx=i+x,ny=j+y; //即代表一个九宫格里面对于中央的W周围的各个位置的方向。if(0<=nx&&nx<N&&0<=ny&&ny<M&&field[nx][ny]=='W') dfs(nx,ny);//如果没有出界而且而且搜索到周围的W，继续dfs}
}
int main()
{int i,j,sum=0;void dfs(int i,int j);  //函数声明scanf("%d %d",&N,&M);for(i=0;i<N;i++){    for(j=0;j<M;j++)scanf(" %c",&field[i][j]);  //输入}for(i=0;i<N;i++){for(j=0;j<M;j++)if(field[i][j]=='W') {dfs(i,j);sum++;}  //找到一个W，则进行dfs}printf("%d\n",sum);return 0;
}

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