HIHO 16 C

树分治。对于一棵子树的根节点，至少有一条边与儿子相连的属于重边。对于一条轻边，它的贡献值是两端子树大小的乘积，所以，重边应该是贡献值最大的一边。

至于要求所有的点，进行深度优先搜索，因为移动一个点只会影响两个点的两个子树，这个可以维护。

在进行DP时，选择计算最大的重边的值，答案就是用所有的边贡献值减去树的重边值的和。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;

const int MAX=100010;

int head[MAX],tol;
struct Edge{
	int u,v,next;
}edge[MAX*2];
int n;

void addedge(int u,int v){
	edge[tol].u=u;
	edge[tol].v=v;
	edge[tol].next=head[u];
	head[u]=tol++;
}

int counts[MAX]; LL dp[MAX];
int weight1[MAX],weight2[MAX];
LL ans[MAX],tot[MAX],par[MAX];

void dfs(int u,int f){
	counts[u]=1;
	weight1[u]=weight2[u]=-1;
	for(int e=head[u];e!=-1;e=edge[e].next){
		int v=edge[e].v;
		if(v!=f){
			dfs(v,u);
			counts[u]+=counts[v];
			dp[u]+=dp[v];
			if(weight1[u]==-1||(LL)(n-counts[v])*counts[v]>(LL)(n-counts[weight1[u]])*counts[weight1[u]]){
				weight2[u]=weight1[u]; weight1[u]=v;
			}
			else if(weight2[u]==-1||(LL)(n-counts[v])*counts[v]>(LL)(n-counts[weight2[u]])*counts[weight2[u]]){
				weight2[u]=v;
			}
	///		dp[u]+=(n-counts[v])*counts[v];
		}
	}
	tot[u]=dp[u];
	if(counts[u]>1){
		dp[u]+=(LL)(n-counts[weight1[u]])*counts[weight1[u]];
	}
}

void dfs_ans(int u,int f){
	for(int e=head[u];e!=-1;e=edge[e].next){
		int v=edge[e].v;
		if(v==f) continue;
		if(v==weight1[u]){
			LL num=max((LL)(n-counts[u])*counts[u],(LL)(n-counts[weight2[u]])*counts[weight2[u]]);
			par[v]=tot[u]-dp[v]+par[u]+num;
		}
		else{
			LL num=max((LL)(n-counts[u])*counts[u],(LL)(n-counts[weight1[u]])*counts[weight1[u]]);
			par[v]=tot[u]-dp[v]+par[u]+num;
		}
		dfs_ans(v,u);
	}
	LL num=max((LL)(n-counts[weight1[u]])*counts[weight1[u]],(LL)(n-counts[u])*counts[u]);
	ans[u]=par[u]+tot[u]+num;
}

int main(){
	while(scanf("%d",&n)!=EOF){
		for(int i=1;i<=n;i++) head[i]=-1,counts[i]=0,dp[i]=0,par[i]=0;
		tol=0;
		int u,v;
		for(int i=1;i<n;i++){
			scanf("%d%d",&u,&v);
			addedge(u,v);
			addedge(v,u);
		}
		dfs(1,0);
		dfs_ans(1,0);
		LL ans_tot=0;
		for(int i=1;i<=n;i++) ans_tot+=(LL)counts[i]*(n-counts[i]);
		for(int i=1;i<=n;i++) cout<<ans_tot-ans[i]<<endl;
	}
	return 0;
}

　　

转载于:https://www.cnblogs.com/jie-dcai/p/4957756.html