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

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.

Input

First 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 payi, ti(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

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”

金策工业综合大学（DPRK）

Source

2016 Multi-University Training Contest 9

#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, -, sizeof d);

    queue<int>q;

    q.push(st);

    d[st] = ;

    while(!q.empty())

    {

        int i,k=q.front();

        q.pop();

        for(i = ; i <= ed; i++)

            if(arc[k][i] >  && d[i] == -)

            {

                d[i] = d[k] + ;

                q.push(i);

            }

    }

    return d[ed]>;

}

int dinic (int k, int low)

{

    if(k == ed)return low;

    int a,i;

    for(i = ; i <= ed; i++)

        if(d[i] == d[k] + &&  arc[k][i] >  &&(a = dinic(i, min(low, arc[k][i]))))

        {

            arc[k][i] -= a;

            arc[i][k] += a;

            return a;

        }

    return ;

}

int cmp(plant a,plant b){

    return a.t<b.t;

}

int main() {

    sf(t);

    for(int cas=;cas<=t;cas++){

        printf("Case #%d: ",cas);

        scanf("%d%d%d",&n,&m,&l);

        for(int i=;i<=n;i++){

            scanf("%d%d",&pt[i].pay,&pt[i].t);

            pt[i].id=i;

        }

        for(int i=;i<=m;i++){

            sf(s[i].pro);

            sf(s[i].k);

            s[i].t=;

            for(int j=;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+,pt++n,cmp);

        int ok=;

        st=n+m+,ed=st+;

        for(int i=;i<=n;i++){

            memset(arc,,sizeof arc);

            for(int j=;j<=i;j++)

                arc[pt[j].id][ed]=pt[j].pay;

            tot=;

            for(int j=;j<=m;j++)if(s[j].t<=pt[i].t){

                tot+=s[j].pro;

                arc[st][j+n]=s[j].pro;

                for(int k=;k<=s[j].k;k++)

                arc[j+n][s[j].pt[k]]=INF;

            }

            ans = ;

            while(bfs())

                while(tans=dinic(st, INF)) ans += tans;

            ans=tot-ans;

            if(ans>=l){

                printf("%d %d\n",pt[i].t,ans);

                ok=;

                break;

            }

        }

        if(!ok)puts("impossible");

    }

}