POJ 1155 树状dp

TELE

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3856		Accepted: 2054

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

Source

POJ 1155

题目意思：一个电视台的信号发射塔和用户以及传输线路之间构成了一个树状结构，根节点代表发射塔，叶子节点代表用户，树的每一条边都有权值代表传输的代价，问从发射塔发射信号，如何在电视台没有损失的情况下使得最多的用户看到电视节目；

解题思路：树状dp,dp[i][j]表示以i为根节点的子树中能满足j个用户的最大余额:

dp[i][j+k]=max(dp[i][j+k],dp[i][j]+dp[son][k]-cost[i][son]);

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
const int maxn=3007;
const int inf=10000000;
struct node
{
    int to,cost;
};
vector<node> G[maxn];
int n,m;
int dp[maxn][maxn];
int value[maxn],num[maxn];
void init()
{
    for(int i=0;i<maxn;i++) G[i].clear();
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
          dp[i][j]=-inf;
    memset(num,0,sizeof(num));
    memset(value,0,sizeof(value));
}
void dfs(int point)
{
    for(int i=0;i<G[point].size();i++)
    {
        int v=G[point][i].to;
        dfs(v);
         for(int j=num[point];j>=0;j--)
            for(int k=1;k<=num[v];k++)
         {
              dp[point][j+k]=max(dp[point][j+k],dp[point][j]+dp[v][k]-G[point][i].cost);
        }
        num[point]+=num[v];
    }
    if(point>=n-m+1) num[point]=1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,a,b;node n1;
    while(~scanf("%d%d",&n,&m)){
        init();
        for(int i=1;i<=n-m;i++)
        {
            scanf("%d",&t);
            for(int j=0;j<t;j++)
            {
                scanf("%d%d",&a,&b);
                n1.to=a;n1.cost=b;
                G[i].push_back(n1);
            }
        }
        for(int i=n-m+1;i<=n;i++)
        {
            scanf("%d",&value[i]);
            dp[i][1]=value[i];
        }
        dfs(1);
        int ans=0;
        for(int i=0;i<=m;i++)
        {
            if(dp[1][i]>=0) ans=i;
        }
        printf("%d
",ans);
    }
    return 0;
}