【洛谷 1596】湖计数

题目描述

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

输入输出格式

输入格式：

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是'W'或'.'，它们表示网格图中的一排。字符之间没有空格。

输出格式：

Line 1: The number of ponds in Farmer John's field.

一行：水坑的数量

输入输出样例

输入样例#1： 复制

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

输出样例#1： 复制

3

说明

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

题解：DFS暴搜石锤

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
bool f[105][105];
//char s[105][105];
string s[105];
int n,m,ans,xx,yy;
int dx[18]={0,1,-1,0,0,1,1,-1,-1};
int dy[18]={0,0,0,1,-1,1,-1,1,-1};
void dfs(int x,int y){
    //if(x>n || x<1 || y>m || y<1) return;
    for(int i=1;i<=8;i++){
        xx=x+dx[i]; yy=y+dy[i];
        //if(f[xx][yy]==1) continue;
        if(s[xx][yy]=='W' && f[xx][yy]==0 &&
           xx<=n && xx>=1 && yy<=m && yy>=1) 
           { f[xx][yy]=1; dfs(xx,yy); } 
    }
}
int main(){
    scanf("%d %d
",&n,&m);
    for(int i=1;i<=n;i++)
        //for(int j=1;j<=m;j++)
            getline(cin,s[i]);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(s[i][j]=='W' && f[i][j]==0) {
                ans++; f[i][j]=1; dfs(i,j);
            }
        }
    }
    cout<<ans;
    return 0;
}