HDOJ 题目4276 The Ghost Blows Light（SPFA+树形DP）

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

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题目大意：一颗树，从1点到n点，在t时间内走到，获得的最大价值

思路：先用spfa求最短路，如果>t,则到不了，t减去最短路，最短路上的边权改为0，如果走的不是最短路的话，来回为两倍的边权，即消耗为两倍的边权

ac代码

#include<stdio.h>
#include<string.h>
#include<queue>
#include<string>
#include<iostream>
#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;
queue<int>q;
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;i<n-1;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
spfa(1);
if(dis
>t)
{
printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
continue;
}
t-=dis
;
dfs(1,-1);
printf("%d\n",dp[1][t]);
}
}