【HDU 5855】Less Time, More profit（网络流、最小割、最大权闭合子图）

饶文津

修改于 2023-09-21 11:16:51

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Less Time, More profit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description

The city planners plan to build N plants in the city which has M shops. Each shop needs products from some plants to make profit of proi units.Building ith plant needs investment of payi units and it takes ti days.Two or more plants can be built simultaneously, so that the time for building multiple plants is maximum of their periods(ti).You should make a plan to make profit of at least L units in the shortest period.InputFirst line contains T, a number of test cases.For each test case, there are three integers N, M, L described above.And there are N lines and each line contains two integers payiti(1<= i <= N).Last there are M lines and for each line, first integer is proi, and there is an integer k and next k integers are index of plants which can produce material to make profit for the shop. 1 <= T <= 30 1 <= N, M <= 200

1≤L,ti≤1000000000

1≤payi,proi≤30000

Output

For each test case, first line contains a line “Case #x: t p”, x is the number of the case, t is the shortest period and p is maximum profit in t hours. You should minimize t first and then maximize p. If this plan is impossible, you should print “Case #x: impossible”

Sample Input

1 1 2

1 5

3 1 1

1 1 3

1 5

3 1 1

Sample Output

Case #1: 5 2

Case #2: impossible

Author

金策工业综合大学（DPRK）

Source

2016 Multi-University Training Contest 9

M个商店，N个工厂，每个商店获利的条件是建设了指定的k个工厂。求总获利不小于L，工厂建设的时间最大值最小是多少。

工厂到汇点建一条边pay[i]，源点到商店建一条边pro[i]，商店到需要的工厂建一条边INF，商店的总收益-最小割就是答案。

可以看看国家集训队论文：Amber《最小割模型在信息学竞赛中的应用》

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 205<<1
#define sf(a) scanf("%d",&a);
using namespace std;
const int INF=0x3f3f3f3f;
struct plant{
    int pay,t,id;
}pt[N];
struct shop{
    int pro,k,pt[N],t;
}s[N];
int t,n,m,l,st,ed,tot;
int arc[N][N], d[N];
int ans,tans;
bool bfs()
{
    memset(d, -1, sizeof d);
    queue<int>q;
    q.push(st);
    d[st] = 0;
    while(!q.empty())
    {
        int i,k=q.front();
        q.pop();
        for(i = 1; i <= ed; i++)
            if(arc[k][i] > 0 && d[i] == -1)
            {
                d[i] = d[k] + 1;
                q.push(i);
            }
    }
    return d[ed]>0;
}
int dinic (int k, int low)
{
    if(k == ed)return low;
    int a,i;
    for(i = 1; i <= ed; i++)
        if(d[i] == d[k] + 1&&  arc[k][i] > 0 &&(a = dinic(i, min(low, arc[k][i]))))
        {
            arc[k][i] -= a;
            arc[i][k] += a;
            return a;
        }
    return 0;
}
int cmp(plant a,plant b){
    return a.t<b.t;
}
int main() {
    sf(t);
    for(int cas=1;cas<=t;cas++){
        printf("Case #%d: ",cas);
        scanf("%d%d%d",&n,&m,&l);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&pt[i].pay,&pt[i].t);
            pt[i].id=i;
        }
        for(int i=1;i<=m;i++){
            sf(s[i].pro);
            sf(s[i].k);
            s[i].t=0;
            for(int j=1;j<=s[i].k;j++){
                int x;
                sf(x);
                s[i].pt[j]=x;
                s[i].t=max(s[i].t,pt[x].t);
            }
        }
        sort(pt+1,pt+1+n,cmp);
        int ok=0;
        st=n+m+1,ed=st+1;
        for(int i=1;i<=n;i++){
            memset(arc,0,sizeof arc);
            for(int j=1;j<=i;j++)
                arc[pt[j].id][ed]=pt[j].pay;
            tot=0;
            for(int j=1;j<=m;j++)if(s[j].t<=pt[i].t){
                tot+=s[j].pro;
                arc[st][j+n]=s[j].pro;
                for(int k=1;k<=s[j].k;k++)
                arc[j+n][s[j].pt[k]]=INF;
            }
            ans = 0;
            while(bfs())
                while(tans=dinic(st, INF)) ans += tans;
            ans=tot-ans;
            if(ans>=l){
                printf("%d %d\n",pt[i].t,ans);
                ok=1;
                break;
            }
        }
        if(!ok)puts("impossible");
    }
}