HDU 4276 The Ghost Blows Light

The Ghost Blows Light

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4276
64-bit integer IO format: %I64d      Java class name: Main

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

5 10
1 2 2 
2 3 2
2 5 3
3 4 3
1 2 3 4 5

Sample Output

11

Source

2012 ACM/ICPC Asia Regional Changchun Online

 

解题：树形dp,反正到n的路上的时间都是必须花的，就把这条路上的时间一并搞掉

然后就是树形dp了

#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
struct arc {
    int to,w,next;
    arc(int x = 0,int y = 0,int z = -1) {
        to = x;
        w = y;
        next = z;
    }
} e[maxn*maxn];
int head[maxn],dp[maxn][510],val[maxn],tot,sum,t,n;
void add(int u,int v,int w) {
    e[tot] = arc(v,w,head[u]);
    head[u] = tot++;
}
bool shortest(int u,int fa) {
    if(u == n) return true;
    for(int i = head[u]; ~i; i = e[i].next) {
        if(e[i].to == fa) continue;
        if(shortest(e[i].to,u)) {
            sum += e[i].w;
            e[i].w = e[i^1].w = 0;
            return true;
        }
    }
    return false;
}
void dfs(int u,int fa){
    for(int i = 0; i <= t; ++i) dp[u][i] = val[u];
    for(int i = head[u]; ~i; i = e[i].next){
        if(e[i].to == fa) continue;
        dfs(e[i].to,u);
        int cost = e[i].w<<1;
        for(int j = t; j >= cost; --j)
            for(int k = 0; k + cost <= j; ++k)
                dp[u][j] = max(dp[u][j],dp[e[i].to][k] + dp[u][j - cost - k]);
    }
}
int main() {
    int u,v,w;
    while(~scanf("%d%d",&n,&t)){
        memset(head,-1,sizeof head);
        sum = tot = 0;
        for(int i = 1; i < n; ++i){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        memset(dp,0,sizeof dp);
        for(int i = 1; i <= n; ++i) scanf("%d",val + i);
        shortest(1,-1);
        if(sum > t) {
            puts("Human beings die in pursuit of wealth, and birds die in pursuit of food!");
            continue;
        }
        t -= sum;
        dfs(1,-1);
        printf("%d
",dp[1][t]);
    }
    return 0;
}