//n件物品,m种关系,(有关系的2个不能在同一组) //把所有物品分为2组,希望最后2组的差值尽可能小,输出较大者 /* 二分图涂色+可行性(01)背包 dp[i] =1表示 最后差值为i可行 建图后,对于每个连通分量记录差值,来求所有的可行 */ #include<bits/stdc++.h> using namespace std; int t,n,m; #define N 250 #define M 102000 int a[N],head[N],sum; int cnt,vis[N]…

Problem Description There are n planets ∼n. Each planet is connected to other planets through some transmission channels. There are m transmission channels in the galaxy. Each transmission channel connects two different planets, and each transmission…

“Miku is matchless in the world!” As everyone knows, Nakano Miku is interested in Japanese generals, so Fuutaro always plays a kind of card game about generals with her. In this game, the players pick up cards with generals, but some generals have co…

Tree Ming and Hong are playing a simple game called nim game. They have nn piles of stones numbered 11 to nn ,the ii-th pile of stones has a_iai​ stones. There are n - 1n−1 bidirectional roads in total. For any two piles, there is a unique path from…

传送门 题意: 给出一个只包含小写字母的串 s 和n 个串t,判断t[i]是否为串 s 的子序列: 如果是,输出"YES",反之,输出"NO": 坑点: 二分一直TLE可还行: 具体思路+细节看代码(有点累了,不想写了) AC代码: #include<iostream> #include<cstdio> #include<vector> #include<cstring> using namespace std; #d…

Travel There are nn planets in the MOT galaxy, and each planet has a unique number from 1 \sim n1∼n. Each planet is connected to other planets through some transmission channels. There are mm transmission channels in the galaxy. Each transmission cha…

Swap There is a sequence of numbers of length nn, and each number in the sequence is different. There are two operations: Swap the first half and the last half of the sequence (if nn is odd, the middle number does not change) Swap all the numbers in…

Angel's Journey “Miyane!” This day Hana asks Miyako for help again. Hana plays the part of angel on the stage show of the cultural festival, and she is going to look for her human friend, Hinata. So she must find the shortest path to Hinata’s house.…

Tasks It's too late now, but you still have too much work to do. There are nn tasks on your list. The ii-th task costs you t_iti​seconds. You want to go to bed TT seconds later. During the TT seconds, you can choose some tasks to do in order to finis…

链接:https://nanti.jisuanke.com/t/39277 思路: 一开始看着很像树分治,就用树分治写了下,发现因为异或操作的特殊性,我们是可以优化树分治中的容斥操作的,不合理的情况只有当两点在一条链上才存在,那么直接一遍dfs从根节点向下跑途中维护一下前缀和,把所有情况中不合理情况造成的值修正. 这样的话时间复杂度就可以降得非常低了,感觉还可以优化,但是懒得写了 代码耗时:142ms. 实现代码: #include<bits/stdc++.h> using namespace…