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…

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:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4738 Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army s…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1231    Accepted Submission(s): 478 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. B…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army still was not good…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 194    Accepted Submission(s): 89 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5050    Accepted Submission(s): 1584 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi.…

http://acm.hdu.edu.cn/showproblem.php?pid=4738 题目大意:曹操有一些岛屿被桥连接,每座都有士兵把守,周瑜想把这些岛屿分成两部分,但他只能炸毁一条桥,问最少需要派几个士兵去;如果不能完成输出-1 1:如果这些岛屿不连通,则不需要派人前去 2:如果桥的守卫是0的话也得派一人去炸毁 3:如果不能完成输出-1 4:输出最少需派的人数 #include<stdio.h> #include<string.h> #include<math.h&…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4738 题目大意:给一些点,用一些边把这些点相连,每一条边上有一个权值.现在要你破坏任意一个边(要付出相应边权值的代价),使得至少有两个连通块.输出最小代价值. 算法思路:这题坑多,要考虑仔细: 1.图是边双连通图,就做不到删除一边得到两个连通块,这种情况输出-1. 2.图是连通但不边双联通,就用tarjan找出桥中权值最小的,这里有个巨坑,如果桥最小的权值为0,这时根据题意,要输出1而不是0(看看题…

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…