HDU 4276 The Ghost Blows Light(树形dp)

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本文解析了一道关于在限定时间内从古墓中获取尽可能多宝藏的算法题。题目涉及图论中的最短路径问题及动态规划算法，通过一次遍历确定最优路径并计算最大收益。

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3844 Accepted Submission(s): 1256

Problem Description

My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road should cost some time). There is exactly one route between any two rooms, and each room contains some treasures. Now I am located at the 1st room and the exit is located at the Nth room.
Suddenly, alert occurred! The tomb will topple down in T minutes, and I should reach exit room in T minutes. Human beings die in pursuit of wealth, and birds die in pursuit of food! Although it is life-threatening time, I also want to get treasure out as much as possible. Now I wonder the maximum number of treasures I can take out in T minutes.

Input

There are multiple test cases.
The first line contains two integer N and T. (1 <= n <= 100, 0 <= T <= 500)
Each of the next N - 1 lines contains three integers a, b, and t indicating there is a road between a and b which costs t minutes. (1<=a<=n, 1<=b<=n, a!=b, 0 <= t <= 100)
The last line contains N integers, which Ai indicating the number of treasure in the ith room. (0 <= Ai <= 100)

Output

For each test case, output an integer indicating the maximum number of treasures I can take out in T minutes; if I cannot get out of the tomb, please output "Human beings die in pursuit of wealth, and birds die in pursuit of food!".

Sample Input

Sample Output

11

题解：
从1到n最短路只有一条，因为这是一棵树。
这条最短路一定只走一次，如果要经过其他
的边，则必定经过两次。
代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
int val[maxn],dp[maxn][5*maxn];
int n,t,d[maxn];
bool is[maxn];
struct node
{
    int to,cost;
};
vector<node>G[maxn];
void bfs()
{
    bool vis[maxn];
    int pre[maxn];
    memset(vis,0,sizeof(vis));
    memset(d,0,sizeof(d));
    queue<int>P;
    P.push(1);vis[1]=1;pre[1]=-1;
    while(!P.empty())
    {
        int v=P.front();P.pop();
        for(int i=0;i<G[v].size();i++)
        {
            node e=G[v][i];
            if(vis[e.to])continue;
            vis[e.to]=1;
            P.push(e.to);
            pre[e.to]=v;
            d[e.to]=d[v]+e.cost;
        }
    }
    int k=n;
    while(k!=-1)
    {
        //printf("%d\n",k);
        is[k]=1;
        k=pre[k];
    }
}
void dfs(int now,int fa)
{
    for(int i=0;i<=t;i++)
        dp[now][i]=val[now];
    for(int i=0;i<G[now].size();i++)
    {
        node e=G[now][i];
        if(e.to==fa)continue;
        dfs(e.to,now);
        for(int j=t;j>=0;j--)
        {
            for(int k=0;k+2*e.cost<=j;k++)
                dp[now][j]=max(dp[now][j],dp[now][j-k-2*e.cost]+dp[e.to][k]);
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&t))
    {
        memset(dp,0,sizeof(dp));
        memset(is,0,sizeof(is));
        for(int i=0;i<=n;i++)
            G[i].clear();
        for(int i=1;i<n;i++)
        {
            int x,y,z;scanf("%d%d%d",&x,&y,&z);
            G[x].push_back((node){y,z});
            G[y].push_back((node){x,z});
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&val[i]);
        bfs();
        if(d[n]>t)
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
        else
        {
            for(int i=1;i<=n;i++)
            {
                if(is[i])
                {
                    for(int j=0;j<G[i].size();j++)
                    {
                        node &e=G[i][j];
                        if(is[e.to])
                            e.cost=0;
                    }
                }
            }
            t-=d[n];
            //printf("%d\n",t);
            dfs(1,-1);
            printf("%d\n",dp[1][t]);
        }
    }
    return 0;
}
/*
4 4
1 2 1
2 4 2
1 3 1
1 1 10 1
*/