Lake Counting【POJ2386】

题目链接

解题思路

1.深搜或者广搜
2.并查集

版本2

这是挑战编程竞赛里的版本，如果数据范围较大，需要使用BFS

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1005;
char g[N][N];
int n, m;int ans = 0;
void dfs(int u, int v){g[u][v] = '.';for(int i = -1; i <= 1; i++ )for(int j = -1; j <= 1; j++){int a = u+i, b = v+j;if(a>=0&&a<n&&b>=0&&b<m&&g[a][b]=='W')  dfs(a, b);}
}
int main(){scanf("%d%d", &n, &m);for(int i = 0; i < n; i++){scanf("%s", g[i]);}for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(g[i][j]=='W') {ans++;dfs(i, j);}}}printf("%d\n", ans);return 0;
}

版本1

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
char arr[105][105];
int x,y;void dfs(int u,int v){if(u<0||u>=x||v<0||v>=y||arr[u][v]=='.') return; //这里要先判断越界与否再判断是否为0arr[u][v]='.';for(int i=-1;i<2;i++){for(int j=-1;j<2;j++){if(i==0&&j==0) continue; else dfs(u+i,v+j);}}
}
int main(){cin>>x>>y;for(int i=0;i<x;i++){for(int j=0;j<y;j++){cin>>arr[i][j];}}int count=0;for(int i=0;i<x;i++){for(int j=0;j<y;j++){if(arr[i][j]=='W'){count++;dfs(i,j);}}}cout<<count<<endl;return 0;
}

Lake Counting【POJ2386】相关推荐

NUC1158 Lake Counting【DFS】
Lake Counting 时间限制: 1000ms 内存限制: 65536KB 通过次数: 1总提交次数: 1 问题描述 Due to recent rains, water has pooled ...
Openjudge1388 Lake Counting【DFS/Flood Fill】
http://blog.csdn.net/c20182030/article/details/52327948 1388:Lake Counting 总时间限制: 1000ms 内存限制: 65 ...
【POJ2386】Lake Counting
problem solution codes //DFS 求通块 #include<iostream> #include<string> using namespace std ...
Triangle Counting【数学】
Triangle Counting UVA - 11401 题目传送门题目大意:输入一个整数n,求在1到n中选取三条边能够组成多少种三角形. AC代码: #include <cstdio> ...
CF990G-GCD Counting【dfs】
正题题目链接:https://www.luogu.com.cn/problem/CF990G 题目大意给出一棵有点权的树,对于每个kkk求有多条路径的点权gcdgcdgcd为kkk 1≤n≤2×1 ...
POJ No.2386【B007】
[B007]Lake Counting[难度B]------------------------------------------ [Description] Due to recent rains ...
【概率论】1-2:计数方法(Counting Methods)
title: [概率论]1-2:计数方法(Counting Methods) categories: Mathematic Probability keywords: Counting Methods ...
【SGU】117. Counting 快速幂
传送门:[SGU]117. Counting 题目分析:快速幂对k取模,结果为0则ans++. 代码如下: #include <map> #include <vector> # ...
DFS:深入优先搜索 POJ-2386 Lake Counting
深度优先搜索是从最开始的状态出发,遍历所有可以到达的状态. 因此可以对所有的状态进行操作,或列举出所有的状态. Lake Counting POJ - 2386 Due to recent rains ...

Lake Counting【POJ2386】

解题思路

版本2

版本1

Lake Counting【POJ2386】相关推荐

最新文章

热门文章