题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4738 题意:有n座岛和m条桥,每条桥上有w个兵守着,现在要派不少于守桥的士兵数的人去炸桥,只能炸一条桥,使得这n座岛不连通,求最少要派多少人去. 分析:只需要用Tarjan算法求出图中权值最小的那条桥就行了.但是这题有神坑. 第一坑:如果图不连通,不用派人去炸桥,直接输出0 第二坑:可能会有重边 第三坑:如果桥上没有士兵守着,那至少要派一个人去炸桥. 比赛的时候看完就想做了,但是图论太挫了,居然不会…

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…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3992    Accepted Submission(s): 1250 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi.…

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…

Burning Bridges Time Limit: 5 Seconds      Memory Limit: 32768 KB Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very beautiful and are loved by everyone in the kingdom. Of course, the sys…

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…

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…

神坑题.这题的坑点有1.判断连通,2.有重边,3.至少要有一个人背*** 因为有重边,tarjan的时候不能用子结点和父节点来判断是不是树边的二次访问,所以我的采用用前向星存边编号的奇偶性关系,用^1来判断是不是树边 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxe = 1000005<<1; const int maxv =…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4738 题意:给定一个n个节点m条边的无向图(可能不连通.有重边),每条边有一个权值.判断其连通性,若双连通,输出-1:若非连通,输出0:否则,输出权值最小的桥的权值. 思路:进行双连通域分解,记下连通块的个数和所有桥的情况,对应输出结果即可. 注意对重边的处理.这里我按照上一道题学到的姿势如法炮制:先把所有边按“字典序”排序(u, v, w),这样重边聚集在一起了,然后扫描一遍,发现重边即在结构体E…

题目链接: Hdu 4738 Caocao's Bridges 题目描述: 有n个岛屿,m个桥,问是否可以去掉一个花费最小的桥,使得岛屿边的不连通? 解题思路: 去掉一个边使得岛屿不连通,那么去掉的这个边一定是一个桥,所以我们只需要求出来所有的桥,然后比较每个桥的花费,选取最小的那个就好. 看起来很简单的样子哦!但是这个题目有很多的细节: A:题目中有重边,以后写Tarjan还是清一色判断重边吧.(除非题目特别要求) B:m个桥有可能连通不了这n个桥,这个时候不需要花费. C:当最小花费桥的花费…