POJ1155 TELE 树形DP

               

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 63 2 2 3 2 9 32 4 2 5 23 6 2 7 2 8 24 3 3 3 1 1

Sample Output

5

题意：电视台发送信号给很多用户，每个用户有愿意出的钱，电视台经过的路线都有一定费用，求电视台不损失的情况下最多给多少用户发送信号。

思路：树状dp。由于求的是最多多少用户，那么我们可以把用户个数当成一个状态。dp[i][j]代表i节点为根节点的子树j个用户的时候最大剩余费用。

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

　　　 注意两点，第一点是上面式子中的dp[i][k]必须先用一个tem[MAX]数组提取出来，因为在计算的过程中会相互影响。第二点是价值可能是负值，所以dp初始化的时候要初始化为负的最大值。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{    int now,val,next;} tree[9005];int n,m,len = 0;int num[3005],head[3005],dp[3005][3005],tem[3005];void add(int i,int x,int y){    tree[len].now = x;    tree[len].val = y;    tree[len].next = head[i];    head[i] = len++;}void dfs(int root){    int i,j,k,p;    for(i = head[root]; i!=-1; i=tree[i].next)    {        p = tree[i].now;        dfs(p);        for(j = 0; j<=num[root]; j++)            tem[j] = dp[root][j];        for(j = 0; j<=num[root]; j++)        {            for(k = 1; k<=num[p]; k++)                dp[root][k+j] = max(dp[root][j+k],tem[j]+dp[p][k]-tree[i].val);        }        num[root]+=num[p];    }}int main(){    int i,j,k,a,b;    while(~scanf("%d%d",&n,&m))    {        memset(head,-1,sizeof(head));        for(i = 1; i<=n-m; i++)        {            scanf("%d",&k);            num[i] = 0;            for(j = 0; j<k; j++)            {                scanf("%d%d",&a,&b);                add(i,a,b);            }        }        for(i = 1; i<=n; i++)        {            for(j = 1; j<=m; j++)                dp[i][j] = -10000000;        }        for(i = n-m+1; i<=n; i++)        {            num[i] = 1;            scanf("%d",&dp[i][1]);        }        dfs(1);        for(i = m; i>=0; i--)        {            if(dp[1][i]>=0)            {                printf("%d\n",i);                break;            }        }    }    return 0;}

           

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