【树形dp】Treasure Hunt I

[ZOJ3626]Treasure Hunt I

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Akiba is a dangerous country since a bloodsucker living there. Sometimes the bloodsucker will appear and kill everyone who isn't at his hometown. One day, a brave person named CC finds a treasure map, and he wants to get as much as possible.

Akiba consists of n towns and n-1 roads. There is a way from each town to any other. Each town contains some treasure values V_i. CC starts from town k(his hometown), at day 0. After m days, the bloodsucker will appear and CC would be killed if he hasn't been back yet, it means CC has m days for hunting the treasure at most. It takes CC T_i days to move from one town to another neighbour town.(Two towns called neighbour if they are the endpoint of one road.) You can assume CC will get the treasure immediately as he arrives at that town. CC wants to obtain as much value as possible, keeping him alive at the same time.

Input

There are multiple cases, about 50 cases.
The first line of each case contains an integer n, indicating there are n towns.
The following line describe the treasure's value in each town. "V₁ V₂ ... V_n". V_i is the value of the treasure in ith town. Each value is separated by one blank.
The next n-1 lines describe the n-1 roads in Akiba. "i j T_i" Means the ith town and the jth town are endpoints of that road. It takes T_i days to get through this road.
The last line has two integer k and m as described above.

1<=n<=100, 0<=V_i<=1000 , 1<=T_i<=10
1<=k<=n, 1<=m<=200
All the inputs are integers.

Output

Just output the max value CC can get, and you should keep CC alive after m days.

Sample Input

2
1 3
1 2 1
1 2
2
1 3
2 1 1
2 1
2
3 3
1 2 1
2 5

Sample Output

4
3
6

Hint

Sample 1: CC can go to town 2 and return at day 2.
Sample 2: CC can't come back within 1 day. So he can only take the treasure in his hometown.
Sample 3: CC only need 2 days to collect all the treasure.

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Author: LI, Chao
Contest: ZOJ Monthly, July 2012

试题分析：设dp[i][j]表示在i号节点走j步能获得的最大宝藏数。

　　　　　就像泛化背包那样，由于还要回去，所以得出转移方程:

　　　　　dp[i][j]=max(dp[i][j-t-2*Cost[x]]+dp[i->son][t],dp[x][j])

　　　　　为什么不是减去2*t呢，因为之前的已经在dp[i->son][t]中算过了。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,M,K;
int val[101],dp[101][201];
int Node[301],Root[301],Cost[301],Next[301];
int cnt;

void addedge(int u,int v,int w){
	cnt++;
	Node[cnt]=v;
	Cost[cnt]=w;
	Next[cnt]=Root[u];
	Root[u]=cnt;
	return ;
}
void init(){
	for(int i=1;i<=N;i++)
	    for(int j=0;j<=M;j++) dp[i][j]=val[i];
	return ;
}
void dfs(int x,int fa){
	for(int i=Root[x];i;i=Next[i]){
		int son=Node[i];
		if(son==fa) continue;
		dfs(son,x);
	}
	for(int i=Root[x];i;i=Next[i]){
		int son=Node[i];
		if(son==fa) continue;
		for(int j=M;j>=1;j--){
			for(int t=0;t<=M;t++){
				if(j<t+2*Cost[i]) break;
				dp[x][j]=max(dp[x][j-t-2*Cost[i]]+dp[son][t],dp[x][j]);
			}
		}
	}
    return ;
}

int main(){
	while(scanf("%d",&N)!=EOF){
		cnt=0;
		memset(Next,0,sizeof(Next));
		memset(Node,0,sizeof(Node));
		memset(Root,0,sizeof(Root));
		memset(Cost,0,sizeof(Cost));
		for(int i=1;i<=N;i++) val[i]=read();
		for(int i=1;i<N;i++){
			int u=read(),v=read(),w=read();
			addedge(u,v,w);
			addedge(v,u,w);
		}
		K=read(),M=read();
		init();
		dfs(K,-1);
		printf("%d\n",dp[K][M]);
	}
}