传送门:http://codeforces.com/contest/1092/problem/F F. Tree with Maximum Cost time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a tree consisting exactly of nn vertices. Tree is a…

题目链接 题意:给你一棵树,让你找一个顶点iii,使得这个点的∑dis(i,j)∗a[j]\sum dis(i,j)*a[j]∑dis(i,j)∗a[j]最大.dis(i,j)dis(i,j)dis(i,j)为iii到jjj的距离. 思路:题解还是好看啊 先从1开始搜,预处理出当1为根的时候的答案记为resresres,递归的过程中假设当前节点为iii,那么sum[i]sum[i]sum[i]数组的意义就是:从以当前顶点为根的所有子树的所有节点的权值和记为sum[i]sum[i]sum[i].…

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…

#include<bits/stdc++.h>using namespace std;const int N=200005;int n,A[N];long long Mx,tot,S[N];vector<int>Adj[N];void DFS(int v,int p){    S[v]=A[v];    for(int &u:Adj[v])        if(u!=p)            DFS(u,v),S[v]+=S[u];    if(p)//0号结点是不存在的…

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传送门 sbsbsb树形dpdpdp题. 题意简述:给出一棵边权为1的树,允许选任意一个点vvv为根,求∑i=1ndist(i,v)∗ai\sum_{i=1}^ndist(i,v)*a_i∑i=1n​dist(i,v)∗ai​的最大值. 直接统计出子树的权值和转移就行了. 代码: #include<bits/stdc++.h> #define ri register int using namespace std; inline int read(){ int ans=0; char ch=g…

Codeforces Round #527 (Div. 3) 题解 题目总链接:https://codeforces.com/contest/1092 A. Uniform String 题意: 输入n,k,n表示字符串的长度,k表示从1-k的小写字符(1即是a),现在要求最大化最少字符的数量. 题解: 贪心搞一搞就行了. 代码如下: #include <bits/stdc++.h> using namespace std; int T; int n,k; int main(){ cin>…

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;…