UVALive - 4614 Moving to Nuremberg （树形DP）

题意：给你一棵树，有一些点你需要去一定的次数，每去一次你都要回来，求最小的权值和

思路：首先将无根树转化话有根树，然后初始化一次求到根的权值和以及每个子树需要去的次数，然后就是怎么求最小的权值和，已知dp[u],怎么求它的子节点v的dp[v],

dp[v] = dp[u] + (sum-f[v])*len - len*f[v],sum表示总的次数,len表示(u,v)的权值的两倍，根据父子节点的关系可以推出来

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 1000000000000000LL
using namespace std;
const int MAXN = 50010;

int n,m,num,cas,u,v;
long long ans,sum,w;
int head[MAXN],next[MAXN*2],ev[MAXN*2],temp[MAXN];
long long fre[MAXN],f[MAXN],ew[MAXN*2],dp[MAXN],dis[MAXN];

void addedge(int u,int v,long long w){
    next[++num] = head[u];
    head[u] = num;
    ev[num] = v;
    ew[num] = w;
}

void read(){
    memset(head,-1,sizeof(head));
    memset(fre,0,sizeof(fre));
    num = -1;
    scanf("%d",&n);
    for (int i = 0; i < n-1; i++){
        scanf("%d%d%lld",&u,&v,&w);
        addedge(u,v,w);
        addedge(v,u,w);
    }
    sum = 0;
    scanf("%d",&m);
    for (int i = 0; i < m; i++){
        scanf("%d%lld",&u,&w);
        fre[u] = w;
        sum += fre[u];
    }
}

void pre_dfs(int fa,int u){
    f[u] = 0;
    dis[u] = 0;
    for (int i = head[u]; i != -1; i = next[i])
        if (ev[i] != fa){
            pre_dfs(u,ev[i]);
            dis[u] += dis[ev[i]] + 2 * ew[i] * f[ev[i]];
            f[u] += f[ev[i]];
        }
    f[u] += fre[u];
}

void dfs(int fa,int u,long long dis){
    dp[u] = dis;
    if (dp[u] < ans)
        ans = dp[u];
    for (int i = head[u]; i != -1; i = next[i])
        if (fa != ev[i])
            dfs(u,ev[i],dis+2*ew[i]*(sum-2*f[ev[i]]));
}

void solve(){
    pre_dfs(1,1);
    ans = INF;
    dfs(1,1,dis[1]);
    printf("%lld\n",ans);
    int cnt = -1;
    for (int i = 1; i <= n; i++)
        if (dp[i] == ans)
            temp[++cnt] = i;
    for (int i = 0; i <= cnt; i++)
        if (i != cnt)
            printf("%d ",temp[i]);
        else printf("%d\n",temp[i]);
}

int main(){
    scanf("%d",&cas);
    while (cas--){
        read();
        solve();
    }
    return 0;
}