Codeforces Round #261 (Div. 2) E - Pashmak and Graph E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pashmak's homework is a problem about graphs. Although he always tr…

E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pashmak's homework is a problem about graphs. Although he always tries to do his homework completely, he can't solve thi…

传送门 解题思路 \(dag\)上\(dp\),首先要按照边权排序,然后图都不用建直接\(dp\)就行了.注意边权相等的要一起处理,具体来讲就是要开一个辅助数组\(g[i]\),来避免同层转移. 代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> using namespace std; const int M…

题目链接:Codeforces 459E Pashmak and Graph 题目大意:给定一张有向图,每条边有它的权值,要求选定一条路线,保证所经过的边权值严格递增,输出最长路径. 解题思路:将边依照权值排序,每次将同样权值的边同一时候增加,维护每一个点作为终止点的最大长度就可以. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn…

E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pashmak's homework is a problem about graphs. Although he always tries to do his homework completely, he can't solve thi…

题目链接: E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pashmak's homework is a problem about graphs. Although he always tries to do his homework completely, he can't sol…

题意:n个点,m条边,每条边有一个权值,找一条边数最多的边权严格递增的路径,输出路径长度. 解法:先将边权从小到大排序,然后从大到小遍历,dp[u]表示从u出发能够构成的严格递增路径的最大长度. dp[u] = max(dp[u],dp[v]+1),因为有重复的边权值,所以用dis数组先记录,到不重复时一起更新重复的那些边权. 代码: (非原创) #include <iostream> #include <cstdio> #include <cstring> #incl…

http://www.codeforces.com/problemset/problem/459/E 题意: 给出n个点,m条边的有向图,每个边有边权,求一条最长的边权上升的路径的长度. 思路:用f存边,g存点,然后排序转移,注意相同的要延迟转移 #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<iostream> struct edge…

http://codeforces.com/contest/459/problem/E 不明确的是我的代码为啥AC不了,我的是记录we[i]以i为结尾的点的最大权值得边,然后wa在第35  36组数据 然后參考答案了,然后----网上一份题解 大意: 给出一个带权有向图,求经过的边权绝对上升的最长路径(可能是非简单路径,就可以能经过一个点多次)所包括的边数. 题解: 对边按权值排序后,从小到大搞. 设q[x]为已经搞过的边组成的以x点为终点的最长路径包括的边数. 设当前边e[i]为从u到v的边,…

题目链接:http://codeforces.com/problemset/problem/459/E 题意: 给你一个有向图,每条边有边权. 让你找出一条路径,使得这条路径上的边权严格递增. 问你这样的路径最长有多长. 题解: 先将所有边按边权从小到大排序,以保证边权递增. 表示状态: dp[i] = max len 表示以点i为终点时的最长路径长度. 找出答案: ans = max dp[i] 如何转移: 枚举每条边e[i],则有: dp[e[i].t] = max(dp[e[i].t],…