在敌人占领之前由城市和公路构成的图是连通图.在敌人占领某个城市之后所有通往这个城市的公路就会被破坏,接下来可能需要修复一些其他被毁坏的公路使得剩下的城市能够互通.修复的代价越大,意味着这个城市越重要.如果剩下的城市无法互通,则说明代价无限大,这个城市至关重要.最后输出的是代价最大的城市序号有序列表.借助并查集和Kruskal算法(最小生成树算法)来解决这个问题. //#include "stdafx.h" #include <iostream> #include <a…

1013 Battle Over Cities (25 分)   It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any o…

1013. Battle Over Cities (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that cit…

1013 Battle Over Cities(25 分) It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any othe…

按题意枚举每个点,建立缺少该点情况下的最小生成树,取权值最大的~ #include<bits/stdc++.h> using namespace std; ; const int inf=1e9; int g[maxn][maxn]; int visit[maxn]; int d[maxn]; int N,M,x,y; int flag; int prim (int s) { fill (d,d+maxn,inf); d[s]=; ; ;i<=N-;i++) { ,min=inf; ;j…

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of…

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of…

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of…

/**题目:删去一个点,然后求出需要增加最小代价的边集合生成连通图思路:prim+最小堆1.之前图中未破坏的边必用,从而把两两之间可互达的点集合 合并成一个点2.求出不同点集合的最短距离,用prim+最小堆求出最小生成树 kruskal1.之前图中未破坏的边必用,全部加到图中2.途中被破坏的边按照边权从小到大的顺序依次加入图中,直到图变为连通图 两个方法的对应一个点的最小生成树的复杂度都是nlogm,第二个方法较好写 优化:1.未破坏的边直接加入图中2.当有n-2条边(当前删去一个点后,图中n-…

题意与分析 题意真的很简单,实在不想讲了,简单说下做法吧. 枚举删除每个点,然后求最小生成树,如果这个路已经存在那么边权就是0,否则按照原来的处理,之后求花费,然后判整个图是否联通(并查集有几个root),如果不联通直接硬点花费是INF,然后处理输出答案即可. 一道最小生成树的模板题,比较有学习的意义. 代码 /* * Filename: pat_top_1001.cpp * Date: 2018-11-05 */ #include <bits/stdc++.h> #define INF 0x…