hdu 4714 树形DP

思路：dp[i][0]表示第i个节点为根的子树变成以i为一头的长链最小的花费，dp[i][0]表示表示第i个节点为根的子树变成i不是头的长链最小花费。

那么动态方程也就不难想了，就是要分几个情况处理，细心就好。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define inf 10000000
#define Maxn 1100010
using namespace std;
int vi[Maxn],head[Maxn],ans,e,dp[Maxn][];
struct Edge{
    int u,v,next,val;
}edge[Maxn*];
void init()
{
    e=;
    ans=;
    memset(head,-,sizeof(head));
    memset(dp,,sizeof(dp));
    memset(vi,,sizeof(vi));
}
void add(int u,int v)
{
    edge[e].u=u,edge[e].v=v,edge[e].next=head[u],head[u]=e++;
    edge[e].v=u,edge[e].u=v,edge[e].next=head[v],head[v]=e++;
}
void dfs(int u)
{
    int i,v;
    vi[u]=;
    int sum1=,num1=,sum2=,num2=,sz=;
    for(i=head[u];i!=-;i=edge[i].next){
        v=edge[i].v;
        if(vi[v]) continue;
        dfs(v);
        if(dp[v][]<=dp[v][]){
            sum1+=dp[v][];
            num1++;
        }else if(dp[v][]<=dp[v][]+){
            sum2+=dp[v][];
            num2++;
        }else {
            sum1+=dp[v][]+;
        }
        sz++;
    }
    //if(u==1) printf("%d %d %d %d\n",num1,sum1,num2,sum2);
    if(sz==) return ;
    if(sz==){
        dp[u][]=sum1+sum2;
        dp[u][]=inf;
        return ;
    }
    if(num1>=){
        dp[u][]+=sum1+sum2+num2+(num1-)*;
        if(num1>=){
            dp[u][]+=sum1+sum2+(num1-)*+num2;
        }
        else {
            dp[u][]+=sum1+sum2+num2;
        }
        return ;
    }
    if(num2>=){
        dp[u][]+=sum1+sum2+(num2-);
        if(num2>=){
            dp[u][]+=sum1+sum2+(num2-);
        }
        else dp[u][]+=sum1+sum2;
        return ;
    }
    dp[u][]+=sum1+sum2;
    dp[u][]+=sum1+sum2;
}
int main()
{
    int t,n,i,j,u,v;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(i=;i<n;i++){
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        dfs();
        //for(i=1;i<=n;i++)
          //  printf("%d %d %d \n",i,dp[i][0],dp[i][1]);
        printf("%d\n",min(dp[][],dp[][])+);
    }
    return ;
}