hdu 4276 树形背包

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1909    Accepted Submission(s): 585

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

5 101 2 2 2 3 22 5 33 4 31 2 3 4 5

 

Sample Output

11

给定一棵n个节点的树，每个节点都有权值，从1点出发，在时间T内到达n，且获得最大的权值和。

由于是树，每两个点之间都有唯一的路径，从1到n之间的必经之路唯一，因此那总时间减去这条路上的权值和，这条路上的边权全部清零，就相当于缩点，

每条边要么访问两次，要么不访问，然后进行一次树形背包得到答案。

代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014-2-6 22:31:52File Name :1.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const int maxn=200;int head[maxn],tol,dp[maxn][5*maxn],weight[maxn],T,t,n;struct node{      int next,to,val;      node(){};      node(int _next,int _to,int _val):next(_next),to(_to),val(_val){}  }edge[5*maxn];  void add(int u,int v,int val){      edge[tol]=node(head[u],v,val);      head[u]=tol++;  }  bool dfs(int u,int pre){    if(u==n)return 1;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(v==pre)continue;if(dfs(v,u)){t+=edge[i].val;edge[i].val=0;return 1;}}return 0;}void dfs1(int u,int fa){      for(int i=0;i<=T;i++)dp[u][i]=weight[u];  for(int i=head[u];i!=-1;i=edge[i].next){  int v=edge[i].to;  if(v==fa)continue;  dfs1(v,u);  int pp=2*edge[i].val;  for(int j=T;j>=pp;j--)  for(int k=0;k<=j-pp;k++)  dp[u][j]=max(dp[u][j],dp[v][k]+dp[u][j-pp-k]);  }}int main(){     //freopen("data.in","r",stdin);     //freopen("data.out","w",stdout);     int i,j,k,p; while(~scanf("%d%d",&n,&T)){ memset(head,-1,sizeof(head));tol=0;         for(i=1;i<n;i++){              scanf("%d%d%d",&j,&k,&p);  add(j,k,p);  add(k,j,p); } for(i=1;i<=n;i++)scanf("%d",&weight[i]); t=0;         dfs(1,1);         if(t>T){ puts("Human beings die in pursuit of wealth, and birds die in pursuit of food!"); continue; } memset(dp,0,sizeof(dp)); T-=t; dfs1(1,-1); cout<<dp[1][T]<<endl; }     return 0;}