Problem 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

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題意：給一棵帶邊權的有根樹，每個葉子節點的權值代表可以補償多少代價，求從根節點最多可以到達多少個葉子，使得付出的總代價不大於0

思路：把每個節點的子節點看作揹包，最大容量是這個點的子孫數量，選幾個結點就是選擇容量，價值就是中轉費用

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 3001
#define MOD 16007
#define E 1e-6
#define LL long long
using namespace std;
struct Edge{
    int to;
    int next;
    int dis;
}edge[N*2];
int n,m;
int head[N],dis[N];
int num[N];
int dp[N][N];
int cnt;
void addEdge(int x,int y,int dis){
    edge[cnt].to=y;
    edge[cnt].next=head[x];
    edge[cnt].dis=dis;
    head[x]=cnt;
    cnt++;
}
void treeDP(int x,int father){
    dp[x][0]=0;
    if(x>n-m){
        dp[x][1]=dis[x];
        num[x]=1;
        return;
    }

    for(int i=head[x];i!=-1;i=edge[i].next){
        int y=edge[i].to;
        if(father==y)//下一個相鄰結點是父節點，到達底層，剪枝
            continue;
        treeDP(y,x);
        num[x]+=num[y];
        for(int j=num[x];j>=0;j--)//容量
            for(int k=1;k<=num[y];k++)//使用者
                dp[x][j]=max(dp[x][j],dp[y][k]+dp[x][j-k]-edge[i].dis);
    }
}
int main()
{
    memset(num,0,sizeof(num));
    memset(dp,~0x3f,sizeof(dp));
    memset(head,-1,sizeof(head));

    scanf("%d%d",&n,&m);
    for(int x=1;x<=n-m;x++){
        int k;
        scanf("%d",&k);
        for(int j=1;j<=k;j++){
            int y,w;
            scanf("%d%d",&y,&w);
            addEdge(x,y,w);
            addEdge(y,x,w);
        }
    }
    for(int i=n-m+1;i<=n;i++)
        scanf("%d",&dis[i]);

    treeDP(1,-1);
    for(int i=n;i>=1;i--){
        if(dp[1][i]>=0){
            printf("%d\n",i);
            break;
        }
    }

	return 0;
}