poj 2152 树形DP

思路：这个没思路，看了陈启峰的论文写得。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<string>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pb push_back
#define mp make_pair
#define Maxn 1010
#define Maxm 80002
#define LL __int64
#define Abs(x) ((x)>0?(x):(-(x)))
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define clr(x,y) memset(x,y,sizeof(x))
#define inf 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define mod 1000000000
using namespace std;
int dp[Maxn][Maxn],best[Maxn],w[Maxn],d[Maxn],n,dis[Maxn],head[Maxn],vi[Maxn],e;
struct Edge{
    int u,v,next,val;
}edge[Maxn*2];
void init()
{
    clr(vi,0);
    clr(head,-1);
    e=0;
}
void add(int u,int v,int val)
{
    edge[e].u=u,edge[e].v=v,edge[e].val=val,edge[e].next=head[u],head[u]=e++;
    edge[e].u=v,edge[e].v=u,edge[e].val=val,edge[e].next=head[v],head[v]=e++;
}
void getdis(int u,int f)
{
    int i,v,sz;
    vi[u]=1;
    for(i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(v==f) continue;
        dis[v]=dis[u]+edge[i].val;
        getdis(v,u);
    }
}
void dfs(int u,int f)
{
    int i,v,j;
    for(i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(v==f) continue;
        dfs(v,u);
    }
    memset(dis,0,sizeof(dis));
    getdis(u,u);
    best[u]=inf;
    for(i=1;i<=n;i++){
        if(dis[i]>d[u]){dp[u][i]=inf;continue;}
        dp[u][i]=w[i];
        for(j=head[u];j!=-1;j=edge[j].next){
            v=edge[j].v;
            if(v==f) continue;
            dp[u][i]+=min(best[v],dp[v][i]-w[i]);
        }
        if(dp[u][i]<best[u])
            best[u]=dp[u][i];
    }
}
int main()
{
    int t,i,j,u,v,val;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",w+i);
        for(i=1;i<=n;i++)
            scanf("%d",d+i);
        for(i=1;i<n;i++){
            scanf("%d%d%d",&u,&v,&val);
            add(u,v,val);
        }
        dfs(1,0);
        printf("%d\n",best[1]);
    }
    return 0;
}