poj1155之树形DP

TELE

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3309		Accepted: 1703

Description

A TV-network plans to broadcast an important football match. Their network of transmitters and users can be represented as a tree. The root of the tree is a transmitter that emits the football match, the leaves of the tree are the potential users and other vertices in the tree are relays (transmitters). 
The price of transmission of a signal from one transmitter to another or to the user is given. A price of the entire broadcast is the sum of prices of all individual signal transmissions. 
Every user is ready to pay a certain amount of money to watch the match and the TV-network then decides whether or not to provide the user with the signal. 
Write a program that will find the maximal number of users able to watch the match so that the TV-network's doesn't lose money from broadcasting the match.

Input

The first line of the input file contains two integers N and M, 2 <= N <= 3000, 1 <= M <= N-1, the number of vertices in the tree and the number of potential users. 
The root of the tree is marked with the number 1, while other transmitters are numbered 2 to N-M and potential users are numbered N-M+1 to N. 
The following N-M lines contain data about the transmitters in the following form: 
K A1 C1 A2 C2 ... AK CK 
Means that a transmitter transmits the signal to K transmitters or users, every one of them described by the pair of numbers A and C, the transmitter or user's number and the cost of transmitting the signal to them. 
The last line contains the data about users, containing M integers representing respectively the price every one of them is willing to pay to watch the match.

Output

The first and the only line of the output file should contain the maximal number of users described in the above text.

Sample Input

9 6
3 2 2 3 2 9 3
2 4 2 5 2
3 6 2 7 2 8 2
4 3 3 3 1 1

Sample Output

5

题目描述：

有一个电视台要用电视网络转播节目。这种电视网络是一树，树的节点为中转站或者用户。树节点的编号为1~N,其中1为总站，2~(N-M)为中转站，（总站和中转站统称为转发站）N-M+1~N为用户，电视节目从一个地方传到另一个地方都要费用，同时每一个用户愿意出相应的钱来付电视节目。现在的问题是，在电视台不亏本的前提下，要你求最多允许有多少个用户可以看到电视节目。

 

输入：

N M N表示转发站和用户总数，M为用户数

以下N-M行，第i行第一个K，表示转发站i和K个（转发站或用户）相连, 其后第j对数val,cost表示，第i个转发站到val有边，费用cost.

最后一行M个数表示每个用户愿意负的钱。

输出：

不亏本前提下，可以收到节目最多的用户数。

（如果某个用户要收到节目（叶子结点），那么电视台到该用户的路径节点的费用都要付）

思路:在树上进行背包，对于以u为根的子树,该子树供给给j个用户亏本的最少钱

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=3000+10;
int head[MAX],size,n,m;
int val[MAX],dp[MAX][MAX];//dp记录以u为根的子树供给j个用户亏损的最少钱

struct Edge{
	int v,w,next;
	Edge(){}
	Edge(int V,int W,int NEXT):v(V),w(W),next(NEXT){}
}edge[MAX*2];

void Init(int num){
	for(int i=1;i<=num;++i){
		head[i]=-1;
		for(int j=1;j<=num;++j)dp[i][j]=INF;
	}
	size=0;
}

void InsertEdge(int u,int v,int w){
	edge[size]=Edge(v,w,head[u]);
	head[u]=size++;
}

int dfs(int u,int father){
	int ans=0,p,now;
	for(int i=head[u];i != -1;i=edge[i].next){
		int v=edge[i].v,w=edge[i].w;
		if(v == father)continue;
		now=dfs(v,u);
		if(v>=n-m+1)++now,p=1;
		else p=0;
		ans+=now;
		for(int j=min(m,ans);j>=1;--j){//这里注意一定要j=min(m,ans),复杂度为O(n^3)=>接近O(n^2)
			for(int k=min(j,now);k>=1;--k){//以u为根的子树供给给j个用户并且有k个用户属于v为根的子树时亏损的最少钱
				dp[u][j]=min(dp[u][j],dp[u][j-k]+dp[v][k-p]+w-val[v]);
			}
		}
	}
	return ans;
}

int main(){
	int k,v,w;
	while(~scanf("%d%d",&n,&m)){
		Init(n);
		for(int i=1;i<=n-m;++i){
			scanf("%d",&k);
			val[i]=0;
			for(int j=1;j<=k;++j){
				scanf("%d%d",&v,&w);
				InsertEdge(i,v,w);
				InsertEdge(v,i,w);
			}
		}
		for(int i=n-m+1;i<=n;++i)scanf("%d",&val[i]);
		dfs(1,-1);
		int sum=0;
		for(n=1;n<=m;++n)if(dp[1][n]<=0)sum=n;
        printf("%d\n",sum);
	}
	return 0;
}