树形DP+背包（poj1155泛化分组背包）

TELE

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3675		Accepted: 1936

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

题意：一个树形网络，编号为1的是广播站，叶子节点为广播接收者，要想使用必须付出费用，中间的其他点是中继站，信号到达每个中继站都会有一定的消费；问在保证广播站收益不亏本的情况下使用者最多是多少？ 分析：dp[i][j]表示第i个点下面有j个使用用户的收益；状态转移方程：                                                       dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]-edge[i].w);                                                             其中u是v的父节点，j枚举u节点的所有子叶点数，k枚举v节点的所有子叶点数，edge[i].w表示u节点下的使用用户不是0的话要减去的成本 dp全部初始化为-inf，dp[i][0]赋为0；

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#define eps 1e-10
#define M 3333
#define inf 99999999
#include"algorithm"
#define g 9.8
#define PI acos(-1.0)
using namespace std;
int dp[M][M];
struct node
{
    int v,w;
    node(int vv,int ww)
    {
        v=vv;
        w=ww;
    }
};
vector<node>edge[M];
int cost[M],sum[M],use[M],m,n;
void dfs(int u)
{
    use[u]=1;
    for(int i=0;i<(int)edge[u].size();i++)
    {
        int v=edge[u][i].v;
        if(!use[v])
        {
            dfs(v);
            sum[u]+=sum[v];
            for(int j=sum[u];j>=0;j--)
            {
                for(int k=1;k<=sum[v];k++)
                {
                    if(j>=k)
                    dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]-edge[u][i].w);
                }
            }
        }
    }
    if(u>n-m)
    {
        sum[u]=1;
        dp[u][1]=cost[u];
    }
}
int main()
{
    int i,j,k,c;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=1;i<=n;i++)
            edge[i].clear();
        for(i=1;i<=n-m;i++)
        {
            scanf("%d",&k);
            while(k--)
            {
                scanf("%d%d",&j,&c);
                edge[i].push_back(node(j,c));
                edge[j].push_back(node(i,c));
            }
        }
        for(i=n-m+1;i<=n;i++)
            scanf("%d",&cost[i]);
        for(i=1;i<=n;i++)
        {
            dp[i][0]=0;
            for(j=1;j<=m;j++)
                dp[i][j]=-inf;
        }
        memset(sum,0,sizeof(sum));
        memset(use,0,sizeof(use));
        dfs(1);
        //for(i=1;i<=m;i++)
            //printf("%d ",dp[1][i]);
        for(i=m;i>=1;i--)
        {
            if(dp[1][i]>=0)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}