洛谷 P1596 [USACO10OCT]湖计数Lake Counting
题目链接
https://www.luogu.org/problemnew/show/P1596
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
思路
没错，我又找到了一个可以暴搜的题目，因为又是判断连通块
似乎并没有什么技术含量的暴搜，有时候，就可以A掉一个题！！
8个if，让你的人生更加美好！！
因为所有的水坑都是连在一起的，所以我们可以for循环找到某一个水坑的某一个元素，将ans加1，然后将其周围的水全部变成陆地（及将这个水坑删除）,然后这样一步一步搜下来就可以找出水坑数啦！
接下来我们就看代码吧！
代码
#include<iostream>
#include<cstdio>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<deque>
using namespace std; int n,m;
char a[][];
bool f[][]= {false}; int baosou(int x,int y) {//将访问的这个水坑删除
if(a[x][y+]=='W') {
a[x][y+]='.';
baosou(x,y+);
}
if(a[x][y-]=='W'){
a[x][y-]='.';
baosou(x,y-);
}
if(a[x+][y]=='W') {
a[x+][y]='.';
baosou(x+,y);
}
if(a[x-][y]=='W') {
a[x-][y]='.';
baosou(x-,y);
}
if(a[x+][y+]=='W') {
a[x+][y+]='.';
baosou(x+,y+);
}
if(a[x+][y-]=='W') {
a[x+][y-]='.';
baosou(x+,y-);
}
if(a[x-][y+]=='W') {
a[x-][y+]='.';
baosou(x-,y+);
}
if(a[x-][y-]=='W') {
a[x-][y-]='.';
baosou(x-,y-);
}
} int main() {
scanf("%d%d",&n,&m);
if(m==)return ;
int ans=;
for(int i=; i<=n; i++) {
for(int j=; j<=m; j++) {
cin>>a[i][j];
//听我的，永远不要用scanf输入字符
}
}
for(int i=; i<=n; i++) {
for(int j=; j<=m; j++) {
if(a[i][j]=='W') {
a[i][j]='.';
ans++;//找到之后将这个点变成旱地并将ans加1
baosou(i,j);//开始暴搜
}
}
}
cout<<ans<<endl;
return ;
}