hdu 4267 树形DP

思路：先dfs一下，找出1，n间的路径长度和价值，回溯时将该路径长度和价值清零。那么对剩下的图就可以直接树形dp求解了。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define inf 100000000
#define Maxn 110
using namespace std;
int vi[Maxn],head[Maxn],dp[Maxn][510],e,n,T,a[Maxn],len,sum;
struct Edge{
    int u,v,next,val;
}edge[Maxn*2];
void init()
{
    memset(vi,0,sizeof(vi));
    memset(head,-1,sizeof(head));
    memset(dp,0,sizeof(dp));
    memset(a,0,sizeof(a));
    e=len=sum=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++;
}
int dfs(int u)
{
    int i,v;
    vi[u]=1;
    if(u==1)
    {
        sum+=a[u];
        a[u]=0;
        return 1;
    }
    for(i=head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(vi[v]) continue;
        if(dfs(v))
        {
            len+=edge[i].val,edge[i].val=edge[i^1].val=0;
            sum+=a[u];
            a[u]=0;
            return 1;
        }
    }
    return 0;
}
void Treedp(int u)
{
    int i,v,j,k;
    vi[u]=1;
    for(i=head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(vi[v]) continue;
        Treedp(v);
        int cost=2*edge[i].val;
        for(j=T;j>=cost;j--){
            int temp=0;
            for(k=0;k<=j-cost;k++)
                temp=max(temp,dp[u][j-cost-k]+dp[v][k]);
                dp[u][j]=max(dp[u][j],temp);
        }

    }
    for(i=0;i<=T;i++)
        dp[u][i]+=a[u];
}
int main()
{
    int i,j,u,v,t;
    while(scanf("%d%d",&n,&T)!=EOF)
    {
        init();
        for(i=1;i<n;i++){
            scanf("%d%d%d",&u,&v,&t);
            add(u,v,t);
        }
        for(i=1;i<=n;i++)
            scanf("%d",a+i);
        dfs(n);
        if(len>T)
        {
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
            continue;
        }
        T-=len;
        memset(vi,0,sizeof(vi));
        Treedp(n);
        printf("%d\n",dp[n][T]+sum);
    }
    return 0;
}