D. Robot Rapping Results Report 题目连接: http://www.codeforces.com/contest/655/problem/D Description While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessi…

题目链接: http://www.codeforces.com/contest/655/problem/D 题意: 题目是要求前k个场次就能确定唯一的拓扑序,求满足条件的最小k. 题解: 二分k的取值,做拓扑排序的时候只要每次只有一个元素没有前驱就可以唯一了. #include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<queue> #includ…

F - Cowslip Collections http://codeforces.com/blog/entry/43868 这个题解讲的很好... #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<LL, int> #define…

E - Intellectual Inquiry 思路:我自己YY了一个算本质不同子序列的方法, 发现和网上都不一样. 我们从每个点出发向其后面第一个a, b, c, d ...连一条边,那么总的不同子序列就是从0号点出发能走出多少条 不同点的路径. dp[ i ]表示是到 i 这个点的不同路径数, 构成的图显然是个DAG,把这个dp出来就好啦.最后补 n个就是贪贪心. 网上的另外两种方法. dp[ i ] 表示[1, i] 的字符串有多少不同子序列. dp[ i ] = dp[i - 1] *…

E. Intellectual Inquiry 题目连接: http://www.codeforces.com/contest/655/problem/E Description After getting kicked out of her reporting job for not knowing the alphabet, Bessie has decided to attend school at the Fillet and Eggs Eater Academy. She has be…

C. Enduring Exodus 题目连接: http://www.codeforces.com/contest/655/problem/C Description In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he…

B. Mischievous Mess Makers 题目连接: http://www.codeforces.com/contest/655/problem/B Description It is a balmy spring afternoon, and Farmer John's n cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through n, are arranged so…

A. Amity Assessment 题目连接: http://www.codeforces.com/contest/655/problem/A Description Bessie the cow and her best friend Elsie each received a sliding puzzle on Pi Day. Their puzzles consist of a 2 × 2 grid and three tiles labeled 'A', 'B', and 'C'.…

题目链接:http://codeforces.com/contest/655/problem/D 大意是给若干对偏序,问最少需要前多少对关系,可以确定所有的大小关系. 解法是二分答案,利用拓扑排序看是否所有关系被唯一确定.即任意一次只能有1个元素入度为0入队. #include <iostream> #include <vector> #include <algorithm> #include <string> #include <string.h&g…

题目链接:https://codeforces.com/contest/1100/problem/E 题意:给出 n 个点 m 条边的有向图,要翻转一些边,使得有向图中不存在环,问翻转的边中最大权值最小是多少. 题解:首先要二分权值,假设当前最大权值是 w,那么大于 w 的边一定不能翻转,所以大于 w 所组成的有向图不能有环,而剩下小于等于 w 的边一定可以构造出一个没有环的图(因为如果一条边正反插入都形成环的话,那么图里之前已经有环了),构造方法是把大于 w 的边跑拓扑序,然后小于等于 w 的…