HDU1811 拓扑排序判环+并查集】的更多相关文章

HDU Rank of Tetris 题目:http://acm.hdu.edu.cn/showproblem.php?pid=1811 题意:中文问题就不解释题意了. 这道题其实就是一个拓扑排序判圈,我的博客里面其他几篇拓扑排序判圈的套路一样.但是这道题与他们不同的的是在大小关系里面存在一种 “=”的关系,这就意味的那些序号不同的点,实际上是一个点.共享入度和出度.我们可以通过并查集将他们合并,合成一个点.这里说一下如何判断信息不完全.我们早先在做拓扑排序,多种排列方式的时候,按照字典序输出.…
Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39602   Accepted: 13944 Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from s…
http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5788    Accepted Submission(s): 2678 Problem Description ACM-DIY is a large QQ group w…
One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my wo…
<题目链接> 题目大意: 给你N个点(编号从0到N-1)和M个关系,要你判断这个图的所有点的顺序是否可以全部确定.不过对于任意点的关系可能存在A>B或A<B或A=B三种情况,如果A=B的话,那么就比较他们的编号,编号大的点分数大.(即两点之间关系如果相等,它们之间的排序也可以确定). 解题分析: 本题要处理等于的情况,单纯的拓扑排序不好处理,  因为这些等于的点与其点的关系是相同的,所以用并查集将这些等于的点进行缩点.当然,在判断冲突的时候,也要考虑到,这些等于的点之间是否会产生冲…
题目: 给出一个无向图,将图中度数小于等于1的点删掉,并删掉与他相连的点,直到不能在删为止,然后判断图中的各个连通分量,如果这个连通分量里边的点的个数是奇数,就把这些点的权值求和. 思路: 先用拓扑排序删点并更新各个点的度数,然后用并查集判断各个连通分量里边的点个数的奇偶性就ok了. 代码: #include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <iostream> #i…
题目描述 Bessie has been playing with strings again. She found that by changing the order of the alphabet she could make some strings come before all the others lexicographically (dictionary ordering). For instance Bessie found that for the strings "omm&…
Almost Acyclic Graph CodeForces - 915D time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a directed graph consisting of n vertices and m edges (each edge is directed, so it can…
Communication 题目链接(点击) 题目描述 The Ministry of Communication has an extremely wonderful message system, designed by the President himself. For maximum efficiency, each office in the Ministry can send a message directly to some, but not necessarily all,…
[题目]D. Almost Acyclic Graph [题意]给定n个点的有向图(无重边),问能否删除一条边使得全图无环.n<=500,m<=10^5. [算法]拓扑排序 [题解]找到一个简单环,则欲删除的边一定经过该环.尝试环上的每一条边(至多n条边)后再次拓扑排序判断全图是否有环. 拓扑排序后定位到简单环:剩余图是环+环内DAG,DFS过程中将走入死路的点标-1,访问过标1,找到访问过的点就是简单环.换起始点直到找到环为止. 复杂度O(nm). #include<cstdio>…