题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6187 题意:有一个V个结点M条边的带边权无向平面图,有一个人在一个区域,要拆一些墙使得他可以到达任意一个区域,问最小花费. 解法: #include <bits/stdc++.h> using namespace std; const int maxn = 100010; const int maxm = 200010; struct edge{ int u,v,w; edge(){} bool…
HDU 6187 Destroy Walls (思维,最大生成树) Destroy Walls *Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1784 Accepted Submission(s): 692 * Problem Description Long times ago, there are beautiful histor…
Destroy Walls Long times ago, there are beautiful historic walls in the city. These walls divide the city into many parts of area. Since it was not convenient, the new king wants to destroy some of these walls, so he can arrive anywhere from his cast…
Destroy Walls Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1774    Accepted Submission(s): 689 Problem Description Long times ago, there are beautiful historic walls in the city. These wal…
Abandoned country 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5723 Description An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) ro…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4940 Destroy Transportation system Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 21    Accepted Submission(s): 17 Problem Description Tom is a…
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5723 [题目大意] n座城市,m条路径,求解: 1.最短的路径和,使得n座城市之间直接或者间接连通 2.在路径和最短的情况下,求出任意两个城市之间的期望距离 [题解] 对于问题1,只需求出该图的最小生成树,边权和即答案,由于边权值唯一,因此不存在最小生成树多解的情况. 对于问题2,期望的通常求法为(任意两点之间的路径和)/(点对数) 那么问题就转化为任意两点间距离和的问题,我们按边考虑,对于每条…
地址:http://acm.hdu.edu.cn/showproblem.php?pid=1301 很明显,这是一道“赤裸裸”的最小生成树的问题: 我这里采用了Kruskal算法,当然用Prim算法也一样可以解题. #include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; typedef struct node{ int f…
Qin Shi Huang's National Road System HDU - 4081 感觉这道题和hdu4756很像... 求最小生成树里面删去一边E1 再加一边E2 求该边两顶点权值和除以(最小生成树-E1)的最大值 其中(最小生成树-E1)必须是最小的 先跑一遍prim 跑完之后在最小生成树里面dp dp[i][j] = i到j的路径中最大的那条边 最小生成树减去dp[i][j]肯定会最小 代码如下 #include <cstdio> #include <algorithm…
Eddy's picture 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 ——每天在线,欢迎留言谈论. 题目大意: 给你N个点,求把这N个点连在一起的最短总距离. 思路: 假设每两两点之间都有路径,求最小生成树. AC代码:(Java) import java.util.Scanner; import java.math.*; public class Main { public static final int MAXN = 110;…