fzu2181(点的双连通分量+求奇环)】的更多相关文章

求出每个点双连通分量,如果在一个点双连通分量中有奇环,则这个分量每个点都在一个奇环中.  关键是要知道怎么求点双连通分量以及点双连通的性质. fzu2181 http://acm.fzu.edu.cn/problem.php?pid=2181 #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define…
题目大意:给出一个 n 点 m 边的图,问最少加多少边使其能够存在奇环,加最少边的情况数有多少种. 解题关键:黑白染色求奇环,利用数量分析求解. 奇环:含有奇数个点的环. 二分图不存在奇环.反之亦成立. #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> using nam…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4612 题目大意:给你一个无向图,问你加一条边后最少还剩下多少多少割边. 解题思路:好水的一道模板题.先缩点变成一颗树,再求树的最长直径,直径两端连一条边就是最优解了. 但是....我WA了一个下午.....没有处理重边. 重边的正确处理方法:只标记已经走过的正反边,而不限制已走过的点.换句话说就是可以经过重边再次走向父亲节点,而不能经过走过边的反向边返回父亲节点. #pragma comment(l…
Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 5093    Accepted Submission(s): 1131 Problem Description N planets are connected by M bidirectional channels that allow instant transport…
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A #include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define pb push_back #define inf 0x3f3f3f3f //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<i…
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By Recognizing These Guys, We Find Social Networks Useful Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 2885    Accepted Submission(s): 726 Problem Description Social Network is popular these…
Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3000    Accepted Submission(s): 953 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. Bu…
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<queue> using namespace std; #define inf 0x3fffffff #define N 1100 #define NN 21000 struct node { int u,v,next; }bian[NN*2]; int head[N],yong; void init() { memset(hea…