poj 2386dfs

时间：2014-11-16 10:36:24 阅读：185 评论：0 收藏：0 [点我收藏+]

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20958		Accepted: 10561

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
bool map[101][101];
char s[101][101];
int ans=0,n,m;
int dic[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
void dfs(int l,int t)
{
     int a,b;
     for(int i=0;i<8;i++)
     {
         a=l+dic[i][0],b=t+dic[i][1];
         if(a>=0&&a<m&&b>=0&&b<n&&map[a][b]==true)
         map[a][b]=false,dfs(a,b);
     }
}
int main()
{
    scanf("%d%d",&m,&n);
    for(int i=0;i<m;i++)
    scanf("%s",s[i]);
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
            {
                if(s[i][j]==‘W‘)
                map[i][j]=true;
                else 
                map[i][j]=false;
            }
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
            if(map[i][j]==true)
               map[i][j]=false,ans++,dfs(i,j);
    printf("%d\n",ans);
    return 0;
}

poj 2386dfs

标签：des blog io ar os sp for strong div

原文地址：http://www.cnblogs.com/a972290869/p/4101053.html

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