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CSP/信奥赛C++刷题训练：经典广搜例题（3）：洛谷P1596 ：[USACO10OCT] Lake Counting S

article 2024/11/8 4:20:43

CSP/信奥赛C++刷题训练：经典广搜例题（3）：洛谷P1596 ：[USACO10OCT] Lake Counting S

在这里插入图片描述

题面翻译

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 $N\times M(1\leq N\leq 100, 1\leq M\leq 100)$ 的网格图表示。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。

输出一行，表示水坑的数量。

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.

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.

输出格式

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

样例 #1

样例输入 #1

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

样例输出 #1

提示

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

使用广搜解题

#include<bits/stdc++.h>
using namespace std;
//bfs：维护一个队列，以一个点往四周搜索，如果符合条件的话就把它放进队列里
int n,m,ans=0;
char a[110][110];
int dx[8]={-1,-1,-1,0,1,1,1,0};//从左上角，顺时针 
int dy[8]={-1,0,1,1,1,0,-1,-1}; 
int q[10010][2];//模拟队列 
void bfs(int x,int y){
	int head=1,tail=1;//头、尾指针
	q[1][0]=x;//当前点入队 
	q[1][1]=y;
	a[x][y]='.';//当前点标记 
	while(head<=tail){
		for(int i=0;i<=7;i++){//搜索当前队首点的8个方向 
			int nx=q[head][0]+dx[i]; 
			int ny=q[head][1]+dy[i];
			if(nx>=1 && nx<=n && ny>=1 && ny<=m && a[nx][ny]=='W'){
				tail++;//队尾下标+1 
				q[tail][0]=nx;//新点入队 
				q[tail][1]=ny; 
				a[nx][ny]='.';//新点标记
			} 
		}
		head++;//删除队首元素 
	} 
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cin>>a[i][j];
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(a[i][j]=='W'){
				bfs(i,j);
				ans++;//答案+1 
			}
		}
	}
	cout<<ans;
	return 0;
}