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  • poj1149 PIGS

    PIGS
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 21917   Accepted: 10034

    Description

    Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs. 
    All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold. 
    More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses. 
    An unlimited number of pigs can be placed in every pig-house. 
    Write a program that will find the maximum number of pigs that he can sell on that day.

    Input

    The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N. 
    The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. 
    The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line): 
    A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

    Output

    The first and only line of the output should contain the number of sold pigs.

    Sample Input

    3 3
    3 1 10
    2 1 2 2
    2 1 3 3
    1 2 6

    Sample Output

    7

    Source

    大致题意:
    M 个猪圈,每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依
    次来了 N 个顾客,每个顾客分别会打开指定的几个猪圈,从中买若干头猪。每
    个顾客分别都有他能够买的数量的上限。每个顾客走后,他打开的那些猪圈中的
    猪,都可以被任意地调换到其它开着的猪圈里,然后所有猪圈重新关上。问总共
    最多能卖出多少头猪。(1 <= N <= 100, 1 <= M <= 1000
    举个例子来说。有 3 个猪圈,初始时分别有 31 10 头猪。依次来了 3 个顾客,
    第一个打开 1 号和 2 号猪圈,最多买 2 头;第二个打开 1 号和 3 号猪圈,最多买
    3 头;第三个打开 2 号猪圈,最多买 6 头。那么,最好的可能性之一就是第一个
    顾客从 1 号圈买 2 头,然后把 1 号圈剩下的 1 头放到 2 号圈;第二个顾客从 3
    号圈买 3 头;第三个顾客从 2 号圈买 2 头。总共卖出 2+3+2=7 头。
    分析:直观地建图,合并点,得到更优的建图方式.具体可以参看Edelweiss的网络流建模汇总这篇文章.
    #include <cstdio>
    #include <queue>
    #include <vector>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    const int maxn = 120,maxm = 2020,inf = 0x7fffffff;
    int S,T,head[maxn],to[maxm],nextt[maxm],tot = 2,c[maxm],w[maxm];
    int n,m,flag[maxm],ans,vis[maxn];
    vector <int> G[maxn];
    
    void add(int x,int y,int z)
    {
        w[tot] = z;
        to[tot] = y;
        nextt[tot] = head[x];
        head[x] = tot++;
    }
    
    bool bfs()
    {
        queue <int> q;
        memset(vis,-1,sizeof(vis));
        vis[S] = 0;
        q.push(S);
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            if (u == T)
                return true;
            for (int i = head[u];i;i = nextt[i])
            {
                int v = to[i];
                if (w[i] && vis[v] == -1)
                {
                    vis[v] = vis[u] + 1;
                    q.push(v);
                }
            }
        }
        return false;
    }
    
    int dfs(int u,int f)
    {
        if (u == T)
            return f;
        int res = 0;
        for (int i = head[u];i;i = nextt[i])
        {
            int v = to[i];
            if (vis[v] == vis[u] + 1 && w[i])
            {
                int temp = dfs(v,min(f - res,w[i]));
                res += temp;
                w[i] -= temp;
                w[i ^ 1] += temp;
                if (res == f)
                    return res;
            }
        }
        if (!res)
            vis[u] = -1;
        return res;
    }
    
    void dinic()
    {
        while (bfs())
            ans += dfs(S,inf);
    }
    
    int main()
    {
        scanf("%d%d",&m,&n);
        S = 0;
        T = n + 1;
        for (int i = 1; i <= m; i++)
            scanf("%d",&c[i]);
        for (int i = 1; i <= n; i++)
        {
            int num;
            scanf("%d",&num);
            for (int j = 1; j <= num; j++)
            {
                int x;
                scanf("%d",&x);
                G[i].push_back(x);
            }
            scanf("%d",&num);
            add(i,T,num);
            add(T,i,0);
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 0;j < G[i].size(); j++)
            {
                int v = G[i][j];
                if (!flag[v])
                {
                    flag[v] = i;
                    add(S,i,c[v]);
                    add(i,S,0);
                }
                else
                {
                    add(flag[v],i,inf);
                    add(i,flag[v],0);
                    flag[v] = i;
                }
            }
        }
        dinic();
        printf("%d
    ",ans);
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zbtrs/p/8137681.html
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