Educational Codeforces Round 67 E.Tree Painting (树形dp)

题目链接

题意:给你一棵无根树，每次你可以选择一个点从白点变成黑点（除第一个点外别的点都要和黑点相邻），变成黑点后可以获得一个权值(白点组成连通块的大小) 问怎么使权值最大

思路:首先，一但根确定了，整棵树的权值就只需要模拟即可，所以思路就转换为求哪一个点为根的权值最大。

这题需要用到一个二次扫描换根的思想,我们可以先从任意一个点去进行树形dp 并且得到从这个点开始去逐渐更新他的儿子节点

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int n;
vector<int> G[200007];
ll dp[200007]; //表示i节点以下的所有贡献
ll f[200007];  //以i为根的权值
ll nump[200007]; //儿子节点数(包含自己)
void dfs(int u,int fa){nump[u]=1;for(int i=0;i<G[u].size();i++){int v=G[u][i];if(v==fa) continue;dfs(v,u);nump[u]+=nump[v];}
}
void dfss(int u,int fa){for(int i=0;i<G[u].size();i++){int v=G[u][i];if(v==fa) continue;dfss(v,u);dp[u]+=dp[v];}dp[u]+=nump[u];
}
void change(int u,int fa){for(int i=0;i<G[u].size();i++){int v=G[u][i];if(v==fa) continue;f[v]=f[u]-nump[v]-dp[v]+n-nump[v]+dp[v]; //核心代码change(v,u);}
}
int main(){ios::sync_with_stdio(false);cin>>n;for(int i=1;i<n;i++){int u,v; cin>>u>>v;G[u].push_back(v);G[v].push_back(u);}dfs(1,0);  dfss(1,0);f[1]=dp[1];change(1,0);ll ans=0;for(int i=1;i<=n;i++){ans=max(ans,f[i]);}cout<<ans<<"\n";return 0;
}

转载于:https://www.cnblogs.com/wmj6/p/11125918.html

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