F. Economic Difficulties 题目链接: https://codeforces.com/contest/1263/problem/F 题目大意: 两棵树,都有n个叶子节点,一棵树正着放,一棵树倒着放,叶子节点从左到右对应装置1,2,3,4...n,问最多能删掉多少条边,使得装置能与两棵树任意一个根节点1相连. 解题思路: mp[ i ][ j ]是装置 i 到 j 这段区间删除这段连续区间所能删除的最大边数,两个图分开看,算出每一个图中如果不连通这段区间对应的叶子节点所能删除…

F. Economic Difficulties An electrical grid in Berland palaces consists of 2 grids: main and reserve. Wires in palaces are made of expensive material, so selling some of them would be a good idea! Each grid (main and reserve) has a head node (its num…

Codeforces Round #485 (Div. 2) F. AND Graph 题目连接: http://codeforces.com/contest/987/problem/F Description You are given a set of size $m$ with integer elements between $0$ and $2^{n}-1$ inclusive. Let's build an undirected graph on these integers in…

Codeforces Round #486 (Div. 3) F. Rain and Umbrellas 题目连接: http://codeforces.com/group/T0ITBvoeEx/contest/988/problem/E Description Polycarp lives on a coordinate line at the point x=0. He goes to his friend that lives at the point x=a. Polycarp can…

题目链接 Codeforces Round #501 (Div. 3) F. Bracket Substring 题解 官方题解 http://codeforces.com/blog/entry/60949 ....看不懂 设dp[i][j][l]表示前i位,左括号-右括号=j,匹配到l了 状态转移,枚举下一个要填的括号,用next数组求状态的l,分别转移 代码 #include<bits/stdc++.h> using namespace std; const int maxn = 207;…

Codeforces Round #499 (Div. 1) F. Tree 题目链接 \(\rm CodeForces\):https://codeforces.com/contest/1010/problem/F Solution 设\(v_i\)表示第\(i\)个点的果子数,设\(b_i=v_i-\sum_{x\in son}v_x\),显然依题意要满足\(b_i\geqslant 0\). 根据差分的性质我们可以得到\(\sum b_i=x\). 假设我们硬点树上剩下了\(m\)个点,则…

Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have three piles of candies: red, green and blue candies: the first pile…

传送门 感觉脑子还是转得太慢了QAQ,一些问题老是想得很慢... A. Sweet Problem 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/29 22:36:19 */ #include <iostream> #include <algorithm> #include <vector> #include <cmath> #include <set> #include <ma…

A. Sweet Problem (找规律) 题目链接 大致思路: 有一点瞎猜的,首先排一个序, \(a_1>a_2>a_3\) ,发现如果 \(a_1>=a_2+a_3\) ,那么答案肯定是 \(a_2+a_3\) ,然后发现发现规律当 \(a_2==a_3\) 的时候,还可以贡献 \((a_1+a_2+a_3)/2\) 的答案,所以只要 \(a_1\) 和 \(a_2\) 减掉一个值就可以了. B. PIN Codes (暴力) 题目链接 大致思路: 可以知道,存在一个数字不同的pi…

题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的倍数,否则就舍弃掉,然后把没有舍弃的数的值加起来,求和的最大值; 43 2 15 9 就拿这个来说,当拿3当做第一个数时结果是3+15+9=27因为2不是3的倍数:当拿2作为第一个数时,结果是2+2+14+8=26因为3,15,9都不是2的倍数,所以只能减小;同理...求最大的和; 我们可以记录每个…