HDU 4276 The Ghost Blows Light

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

解题思路：由题意我们知道从1到n的路径上的边仅经过一次，其余的边要么不经过要么经过两次，因此我们可以首先将1->n路径上的边的权值设为0，这样我们利用分组背包的树形DP进行求解时是可以保证这条路径上的边肯定经过且仅经过一次。

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <numeric>#include <algorithm>#include <functional>using namespace std;const int maxn = 110;const int inf  = 0x3f3f3f3f;int n, T, u, v, w;struct Edge {    int u, v, w, next;    Edge() { }    Edge(int _u, int _v, int _w, int _next) : u(_u), v(_v), w(_w), next(_next) { }}edges[2*maxn];int head[maxn], edge_sum;void init_graph() {    edge_sum = 0;    memset(head, -1, sizeof(head));}void add_edge(int u, int v, int w) {    edges[edge_sum].u = u;    edges[edge_sum].v = v;    edges[edge_sum].w = w;    edges[edge_sum].next = head[u];    head[u] = edge_sum++;    edges[edge_sum].u = v;    edges[edge_sum].v = u;    edges[edge_sum].w = w;    edges[edge_sum].next = head[v];    head[v] = edge_sum++;}int node[maxn];int pre[2*maxn];int dp[maxn][510];void dfs1(int u, int fa) {    for(int i = head[u]; i != -1; i = edges[i].next) {        int v = edges[i].v;        if(v == fa) continue;        pre[v] = i;        dfs1(v, u);    }    return ;}void dfs2(int u, int fa) {    for(int i = 0; i <= T; ++i) {        dp[u][i] = node[u];    }    for(int i = head[u]; i != -1; i = edges[i].next) {        int v = edges[i].v;        int w = edges[i].w;        if(v == fa) continue;        dfs2(v, u);        for(int j = T; j >= 0; --j) {            for(int k = 0; k + 2*w <= j; ++k) {                dp[u][j] = max(dp[u][j], dp[u][j-k-2*w] + dp[v][k]);            }        }    }    return ;}void solve() {    int tot_w = 0;    memset(pre, -1, sizeof(pre));    dfs1(1, -1);    int id = pre[n];    while(id != -1) {        tot_w += edges[id].w;        edges[id].w = 0;        edges[id^1].w = 0;        id = pre[edges[id].u];    }    if(tot_w > T) {        printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");    } else {        T -= tot_w;        dfs2(1, -1);        printf("%d\n", dp[1][T]);    }    return ;}int main() {    //freopen("aa.in", "r", stdin);    while(scanf("%d %d", &n, &T) != EOF) {        init_graph();        for(int i = 1; i < n; ++i) {            scanf("%d %d %d", &u, &v, &w);            add_edge(u, v, w);        }        for(int i = 1; i <= n; ++i) {            scanf("%d", &node[i]);        }        solve();    }    return 0;}