The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2549 Accepted Submission(s): 795


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 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
Recommend liuyiding | We have carefully selected several similar problems for you: 4272 4277 4274 4269 4275 題目大意：一顆樹，從1點到n點，在t時間內走到，獲得的最大價值 思路：先用spfa求最短路，如果>t,則到不了，t減去最短路，最短路上的邊權改為0，如果走的不是最短路的話，來回為兩倍的邊權，即消耗為兩倍的邊權 ac代碼

#include
#include
#include
#include
#include
#define max(a,b) (a>b?a:b)
#define INF 0xfffffff
using namespace std;
int n,t;
int a[110],head[110],vis[110],cnt,pre[110],dis[110],d[220],dp[110][10100];
struct s
{
	int u,v,w,next;
}edge[220];
void add(int u,int v,int w)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].w=w;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void spfa(int u)
{
	int i;
	for(i=1;i<=n;i++)
	{
		dis[i]=INF;
		pre[i]=i;
		d[i]=0;
	}
	memset(vis,0,sizeof(vis));
	vis[u]=1;
	dis[u]=0;
	queueq;
	q.push(u);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].v;
			int w=edge[i].w;
			if(dis[v]>dis[u]+w)
			{
				dis[v]=dis[u]+w;
				pre[v]=u;
				d[v]=i;
				if(!vis[v])
				{
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
	for(i=n;i!=1;i=pre[i])
	{
		edge[d[i]].w=0;
		edge[d[i]^1].w=0;
	}
}
void dfs(int u,int pre)
{
	int i,j,k;
	for(i=0;i<=t;i++)
	{
		dp[u][i]=a[u];
	}
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(v==pre)
			continue;
		dfs(v,u);
		int cost=edge[i].w*2;
		for(j=t;j>=cost;j--)
		{
			for(k=0;k+cost<=j;k++)
			{
				dp[u][j]=max(dp[u][j],dp[v][k]+dp[u][j-k-cost]);
			}
		}
	}
}
int main()
{
	while(scanf(%d%d,&n,&t)!=EOF)
	{
		int i;
		memset(head,-1,sizeof(head));
		memset(dp,0,sizeof(dp));
		cnt=0;
		for(i=0;it)
		{
			printf(Human beings die in pursuit of wealth, and birds die in pursuit of food!
);
			continue;
		}
		t-=dis[n];
		dfs(1,-1);
		printf(%d
,dp[1][t]);
	}
}