文章标题 POJ 2386 : Lake Counting（BFS）

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本文介绍了一道典型的图遍历问题——湖泊计数。通过使用广度优先搜索(BFS)算法，解决如何计算地图上由'W'字符表示的水区域数量。给出详细实现代码，帮助读者理解算法原理及应用场景。

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Lake Counting

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

USACO 2004 November
题意：由于下雨，一些地方有积水，求一个区域中有多少块积水区域。给你一张图，上面有水的地方就是W，他的周围8个方向的W也是同一个区域，求总共有多少个区域。
分析：就是一道BFS的题，求连通块。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
char map1[105][105];
int n,m;
struct node {
    int x,y;
};
int dx[8]={1,1,0,-1,-1,-1,0,1};
int dy[8]={0,1,1,1,0,-1,-1,-1};
void bfs(int x,int y){
    queue<node>q;
    node t;
    t.x = x;
    t.y = y;
    q.push(t);
    while (!q.empty()){
        node temp;
        temp=q.front() ;
        q.pop() ;
        int xn;
        int yn;
        for (int i=0;i<8;i++){
            xn=temp.x+dx[i],yn=temp.y+dy[i];
            if (xn>=0 && xn<n && yn>=0 && yn<m && map1[xn][yn]=='W'){
                map1[xn][yn]='.';
                node a;
                a.x=xn,a.y=yn;
                q.push(a); 
            }
        } 
    } 
} 
int main ()
{
    while (scanf ("%d%d",&n,&m)!=EOF){
        for (int i=0;i<n;i++){
            scanf ("%s",map1[i]);
        }
        int cnt=0;
        for (int i=0;i<n;i++){
            for (int j=0;j<m;j++){
                if (map1[i][j]=='W'){
                    bfs(i,j);
                    map1[i][j]='.';
                    cnt++;
                }
            }
        }
        printf ("%d\n",cnt);
    }       
    return 0;
}