题目链接: Hdu 4612 Warm up 题目描述: 给一个无向连通图,问加上一条边后,桥的数目最少会有几个? 解题思路: 题目描述很清楚,题目也很裸,就是一眼看穿怎么做的,先求出来双连通分量,然后缩点重新建图,用bfs求树的直径,直径的长度就是减去桥的数目. 这个题目需要手动扩展,而且手动扩展的话要用C++提交,G++re哭了. #include <cstdio> #include <queue> #include <cstring> #include <i…

Warm up Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4612 Description N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel bet…

Warm up 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4612 Description N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels. If we can i…

http://acm.hdu.edu.cn/showproblem.php?pid=4612 将原图进行缩点 变成一个树 树上每条边都是一个桥 然后加一条边要加在树的直径两端才最优 代码: #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cmath> #include<set> #include<map> #in…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4612 题意:一个包含n个节点m条边的无向连通图(无自环,可能有重边).求添加一条边后最少剩余的桥的数目. 思路:要想尽可能地消灭桥,那么添加的这条边一定是连通了最多的BCC. 所以首先进行双连通分量分解,并记录桥的数目:然后将同属一个BCC的点缩成一个,代之以block序号,以block序号为点将原图重构为一棵树. 最后求树的直径,桥的数目减去树的直径即为答案. 整体框架是学习了 http://w…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 3160    Accepted Submission(s): 718 Problem Description N planets are connected by M bidirectional channels that allow instant transport…

Warm up Problem Description   N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels. If we can isolate some planets from others by breaking…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 90    Accepted Submission(s): 12 Problem Description N planets are connected by M bidirectional channels that allow instant transportatio…

4612Warm hdu up 题目:给出一个图,添加一条边之后,问能够在新图中得到的最少的桥的数量. 分析:我们可以双联通分量进行缩点,原图变成了一棵树.问题变成了:求树中添加一条边之后,使得不在圈的边最少.显然求一边直径,用总边数减掉最长路上的边数就是答案.注意数据存在重边的情况. #include <set> #include <map> #include <list> #include <cmath> #include <queue> #…

本来就是自己负责图论,结果水了= = 题目其实很裸,就是求桥的数量,只是要新加上一条边罢了.做法:先缩点.再在树上搜最长链(第一场多校的hdu 4607Park Visit就考了最长链,小样,套个马甲以为就认不出你了),加边后求桥数就可以了. 犯了一大三小四个错误-_- 竟然真的去套模板求桥数了..竟然没注意到树边都是桥,用树边减链边就可以了. 然后,重新建图的Build()把n传进去了. 再然后,发现读数据从0~m-1,Build()里竟然是1~m. 再再然后,原来存边的su[],sv[]数组…