HDU 4276 The Ghost Blows Light 第37届ACM/ICPC长春赛区1010题 （树形DP）

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 590 Accepted Submission(s): 194


[align=left]Problem Description[/align]

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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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!".

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

11

[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Changchun Online

[align=left]Recommend[/align]
liuyiding

比较典型的树形DP题。比赛时没有做出来，还要加强下树形DP；
思路就是先找出1到N的路径，然后将这些路径的时间置0，之后进行树形DP

/*
树形DP
先找出1到N的边，将这些边的时间修改为0.然后就是简单的树形DP了
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=110;

struct Node
{
int next;
int to;
int val;
}edge[MAXN*2];
int tol;
int head[MAXN];

int dp[MAXN][550];//dp[i][j]表示从第i个点开始，回到i点，花费j时间得到的最大财富值

int value[MAXN];//每个点的财富值

int time1;//从1到N需要的时间
int n;

void init()
{
tol=0;
memset(head,-1,sizeof(head));
}

void add(int a,int b,int val)
{
edge[tol].to=b;
edge[tol].next=head[a];
edge[tol].val=val;
head[a]=tol++;
edge[tol].to=a;
edge[tol].next=head[b];
edge[tol].val=val;
head[b]=tol++;
}

bool dfs1(int u,int pre)
{
if(u==n)return true;//找到了
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v==pre)continue;
if(dfs1(v,u))
{
time1+=edge[i].val;
edge[i].val=0;
return true;
}
}
return false;
}
int t;
void dfs2(int u,int pre)
{
for(int i=0;i<=t;i++) dp[u][i]=value[u];
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v==pre)continue;
dfs2(v,u);
int cost=edge[i].val*2;//要走两遍
for(int i=t;i>=cost;i--)
for(int j=0;j<=i-cost;j++)
dp[u][i]=max(dp[u][i],dp[v][j]+dp[u][i-j-cost]);
}
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int u,v,w;
while(scanf("%d%d",&n,&t)!=EOF)
{
init();
for(int i=1;i<n;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
for(int i=1;i<=n;i++)scanf("%d",&value[i]);
time1=0;
dfs1(1,-1);//找从1到N的最短时间
if(t<time1)
{
printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
continue;
}
t-=time1;
dfs2(1,-1);
printf("%d\n",dp[1][t]);
}
return 0;
}