Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 49820 | Accepted: 24452 |
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.
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
题目大意：
最近总是下雨、约翰想统计院子里面有多少处连着的水洼，" W "代表水洼、" . "代表陆地，W只要连着就行，就是一个大水洼。题目要求统计共有几个大水洼。
解题思路：
遍历整个图，只要遇到“ W ”就深搜，八个方向遇到“ W ”就变成“ . ”，运行几次深搜，就是题目的答案。
AC代码如下：
#include<iostream>
using namespace std;
//W .
int N,M;
//N行M列
char mapp[101][101];
int nextt[8][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};
void dfs(int x,int y)
{
mapp[x][y]='.';
for(int i=0;i<8;i++)
{
int gx=x+nextt[i][0];
int gy=y+nextt[i][1];
if(gx>=0&&gx<N&&gy>=0&&gy<M&&mapp[gx][gy]=='W')
dfs(gx,gy);
}
}
int main()
{
int ans=0;
cin >> N >> M;
for(int i=0;i<N;i++)
for(int j=0;j<M;j++)
cin >> mapp[i][j];
for(int i=0;i<N;i++)
{
for(int j=0;j<M;j++)
{
if(mapp[i][j]=='W')
{
dfs(i,j);
ans++;
}
}
}
cout << ans << endl;
return 0;
}