题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4587 题意: 删除两个点,使连通块的数目最大化 题解: 枚举删除第一个点,然后对删除了第一个点的图跑割点更新答案. 代码: #include<algorithm> #include<iostream> #include<cstring> #include<vector> #include<cstdio> using namespace std; ;…

问一个无向图中去掉任意两点后剩下的连通分量的个数最大值 枚举第一个删去的点,在剩下的子图中求割点 注意,剩下的子图可能不连通,那么就要对每个连通块求割点 计算删去一个点后剩余连通分量个数 left 的方法为:tarjan算法中的时间戳数组dfn[]若为0说明是新的连通分量 求删去割点后剩余连通分量个数: tarjan算法中将判断是否为割点的bool 数组改为int类型,并将iscut[i] = 1 改为 iscut[i]++ 即可 那么对于非根节点,删去后剩余个数为iscut[i] + 1(子树…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4587 题目给了12000ms,对于tarjan这种O(|V|+|E|)复杂度的算法来说,暴力是能狗住的.可以对每个点进行枚举,然后对剩余的网络进行tarjan,对割点所能造成的最大的连通分量进行查询,也就是如下的方程.ans=max{cut[i]}+cnt 其中cnt删除第一个结点之后剩下的网络在初始时刻的连通分量的数量,也就是对每一个第一结点tarjan进行深搜的次数.另外,这次的tarjan中的…

POJ1144 网络 description: 给出一张\(N\)个点的无向图,求其中割点的个数 data range: \(N\le 100\) solution: 一道模板题(但是读入实在是把我恶心坏了) #include<cstdio> #include<vector> #include<algorithm> #include<iostream> #include<sstream> using namespace std; const in…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4587 TWO NODES Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1448    Accepted Submission(s): 441 Problem Description Suppose that G is an undir…

TWO NODES Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2072    Accepted Submission(s): 683 Problem Description Suppose that G is an undirected graph, and the value of stab is defined as fol…

Suppose that G is an undirected graph, and the value of stab is defined as follows: Among the expression,G -i, -j is the remainder after removing node i, node j and all edges that are directly relevant to the previous two nodes. cntCompent is the num…

#include <bits/stdc++.h> using namespace std; ; ; struct Edge { int to, next; } edge[M]; int head[N]; int cntE; void addedge(int u, int v) { edge[cntE].to = v; edge[cntE].next = head[u]; head[u] = cntE++; edge[cntE].to = u; edge[cntE].next = head[v]…

Description Suppose that G is an undirected graph, and the value of stab is defined as follows: Among the expression,G -i, -j is the remainder after removing node i, node j and all edges that are directly relevant to the previous two nodes. cntCompen…

SPF Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8139   Accepted: 3723 Description Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a…