Travel Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1313    Accepted Submission(s): 472 Problem Description Jack likes to travel around the world, but he doesn’t like to wait. Now, he is tr…

Travel Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5441 Description Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are n cities and m…

Time Limit: 1500/1000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Description Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are n cities and m bidir…

Problem Description TT and FF are ... friends. Uh... very very good friends -________-bFF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integer…

http://acm.hdu.edu.cn/showproblem.php?pid=5441 题意:给出一个图,每条边有一个距离,现在有多个询问,每个询问有一个距离值d,对于每一个询问,计算出有多少点对(x,y)使得在x到y的路径上没有一条边的距离大于d. 思路:只要边距离小于d,那就是可行的,直接加入并查集来维护.并查集需要维护一下树的节点个数. 将边和询问都排序离线处理. #include<iostream> #include<cstdio> #include<cstri…

题意:给你一个带权的无向图,然后q(q≤5000)次询问,问有多少对城市(城市对(u,v)与(v,u)算不同的城市对,而且u≠v)之间的边的长度不超过d(如果城市u到城市v途经城市w, 那么需要城市u到城市w的长度e1≤d,同时城市w到城市v的长度e2≤d). 析:一开始的时候,题意都读错了,怎么看都不对,原来是只要最大的d小于等于x就可以,过了好几天才知道是这样..... 这个题是并查集的应用,先从d小的开始遍历,然后去判断有几个连通块,在连通块中计数,用一个数组即可,运用排列组合的知识可以知…

Problem Description With the 60th anniversary celebration of Nanjing University of Science and Technology coming soon, the university sets n tourist spots to welcome guests. Of course, Redwood forests in our university and its Orychophragmus violaceu…

India and China Origins Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description A long time ago there are no himalayas between India and China, the both cultures are frequently exchanged and are kept in…

题目链接 一开始是想不断的把边插进去,然后再去考虑我们每次都加进去边权为1的边,直到跑到第几次就没法继续跑下去的这样的思路,果不其然的T了. 然后,就是想办法咯,就想到了二分答案. 首先,我们一开始处理关系,(一开始看错了男女关系,结局懵逼了好久),注意输入是女选男.然后,就是去处理咯,我们先要去考虑,女生之间为朋友的话,又由于朋友关系是可以推的,所以我们不妨用并查集去维护这层关系,并且把总的关系推上到并查集的根上去. 然后,就是怎么样去想这个二分答案的过程了,我们可以假设玩了x轮,这么就代表了…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4750 题目大意: 给一无向图,n个点,m条边,每条边有个长度,且不一样.定义f(i,j)表示从节点i到节点j的所有路径中的最大边权值的最小值.有q个询问,每个询问有个t,求f(i,j)>=t的种数. 解题思路: 并查集+简单dp+二分. 比赛的时候各种TLE和MLE.只是查找方式不对. 队友思路,先按边从小到大排序考虑,对于每条边E该边两个节点为a.b,如果a.b不在同一个联通块,则a联通块中点集…