Avoid The Lakes Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6826   Accepted: 3637 Description Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of wate…

Avoid The Lakes Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6795   Accepted: 3622 Description Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of wate…

Avoid The Lakes Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8173   Accepted: 4270 Description Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of wate…

题意:给出一个农田的图,n行m列,再给出k个被淹没的坐标( i , j ).求出其中相连的被淹没的农田的最大范围. 思路:dfs算法 代码: #include<iostream> #include<stdio.h> using namespace std; int t[150][150]; int visit[150][150]; int n,m,k; int dfs(int i,int j){ if(i<1||j<1||i>n||j>m||t[i][j]=…

Description Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the larges…

http://poj.org/problem?id=3620 DFS 从任意一个lake出发 重置联通的lake 并且记录 更新ans #include <iostream> #include <string.h> #include <stdio.h> using namespace std; int N,M,K; ][]; ][] = {-, , , , , , , -}; ; ; bool OK(int x, int y) { || x > N || y &l…

题目 找最大的一片湖的面积,4便有1边相连算相连,4角不算. runtime error 有一种可能是 数组开的太小,越界了 #define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<queue> using namespace std; #define MAXN 110 in…

一.题目 输入一个m行n列的字符矩阵,统计字符“@”组成多少个八连块.如果两个字符所在的格子相邻(横.竖.或者对角线方向),就说它们属于同一个八连块. 二.解题思路 和前面的二叉树遍历类似,图也有DFS和BFS遍历.由于DFS更容易编写,一般用DFS找联通块:从每个“@”格子出发,递归遍历与之相邻的“@”格子,标记相同的“联通分量标号”.这样在访问之前需检查它是否已经有了编号,从而避免一个格子访问多次. 三.代码实现 #include<stdio.h> #include<iostream…

/*572 - Oil Deposits ---DFS求联通块个数:从每个@出发遍历它周围的@.每次访问一个格子就给它一个联通编号,在访问之前,先检查他是否 ---已有编号,从而避免了一个格子重复访问多次 --*/ #define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<string.h> #include<algorithm> using namespace std; const int maxn =…

一.题目(CF 598D) 输入一个n x m的字符矩阵,求从某个空点出发,能碰到多少面墙壁,总共询问k次.(3 ≤m,n ≤1000,1 ≤ k ≤ min(nm,100 000)) 二.解题思路 用DFS找连通分量:从每个“.”格子出发,递归遍历与之相邻的“*”格子,且写上相同的联通分量(即代码中的blocks数组),同时统计该联通分量边界墙壁面数.由于存在多次查询,我们用blocks[i][j]记录格子(i,j)所在的联通分量标号,用res[i]表示联通分量i的边界墙壁面数. 三.代码实现…