zoukankan      html  css  js  c++  java
  • 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
    ",i);
                    break;
                }
            }
        }
        return 0;
    }
    


  • 相关阅读:
    Angular指令2
    MD测试
    LeeCX
    Java公众号推荐
    基于云落地SLB+Tengine实现高可用集群负载均衡
    非HTTP状态下开启web视频的临时方案
    慕课网 20200330 es+geo+baidu-map 直播视频与文字版笔记
    [涨姿势]Chrome 谷歌浏览器原生截长图命令
    人人都要关注的分布式系统中的CAP理论
    疫情风险下的项目管理与紧急预案策略
  • 原文地址:https://www.cnblogs.com/mypsq/p/4348166.html
Copyright © 2011-2022 走看看