DZY Loves Topological Sorting Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5195 Description A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for ev…
传送门 DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 52 Problem Description A topological sort or topological ordering of a directed…
DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1250    Accepted Submission(s): 403 Problem Description A topological sort or topological ordering of a directed g…
DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 866    Accepted Submission(s): 250 Problem Description A topological sort or topological ordering of a directed g…
题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5195 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=573&pid=1002 题解: 1.拓扑排序+贪心 #include<algorithm> #include<iostream> #include<cstring> #include<cstdi…
题意: 删去K条边,使拓扑排序后序列字典序最大 分析: 因为我们要求最后的拓扑序列字典序最大,所以一定要贪心地将标号越大的点越早入队.我们定义点i的入度为di. 假设当前还能删去k条边,那么我们一定会把当前还没入队的di≤k的最大的i找出来,把它的di条入边都删掉,然后加入拓扑序列. 删除的一定是小连大的边,因为大连小的边在拓扑序列生成的时候就去掉了 #include <iostream> #include <cstdio> #include <queue> #incl…
Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…
传送门 题意简述:给出一张DAGDAGDAG,要求删去不超过kkk条边问最后拓扑序的最大字典序是多少. 思路:贪心帮当前不超过删边上限且权值最大的点删边,用线段树维护一下每个点的入度来支持查询即可. 注意要在选点的时候更新后继的入度 代码: #include<bits/stdc++.h> #define ri register int using namespace std; inline int read(){ int ans=0; char ch=getchar(); while(!isd…
Description In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2). DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: …
题目地址:HDU 5266 这题用转RMQ求LCA的方法来做的很easy,仅仅须要找到l-r区间内的dfs序最大的和最小的就能够.那么用线段树或者RMQ维护一下区间最值就能够了.然后就是找dfs序最大的点和dfs序最小的点的近期公共祖先了. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm>…