HDU 5855Less Time, More profit

Less Time, More profit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 468    Accepted Submission(s): 172

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

 

 

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

2

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）

最大权闭合图。源点与利润x建边，边权为利润值。花费y与汇建边，权值为花费值。  x与y之间的依赖关系 建边，权值为 无穷。期望总利润-最小割（最大流）就是能获得的最大利润。

本题因为有时间限制，所以加个二分。在满足利润>=L的情况下求最小时间。在最小时间下求最大利润。

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/17 13:55:10
File Name     :hdu5855.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
struct Edge{
    int from,to,cap,flow;
};
struct node{
    int n,m,s,t;//节点数 边数 源点 汇点
    vector<Edge>edges;
    vector<int>G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void init(int n,int s,int t){
        this->n=n;
        this->s=s;
        this->t=t;
        for(int i=0;i<n;i++)G[i].clear();
        edges.clear();
        m=0;
    }
    void addEdge(int from,int to,int cap){
        edges.push_back({from,to,cap,0});
        edges.push_back({to,from,0,0});
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BFS(){
        memset(vis,0,sizeof vis);
        queue<int>que;
        que.push(s);
        d[s]=0;
        vis[s]=true;
        while(!que.empty()){
            int x=que.front();que.pop();
            for(int i=0;i<G[x].size();i++){
                Edge&e=edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=true;
                    d[e.to]=d[x]+1;
                    que.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x,int a){
        if(x==t||a==0)return a;
        int flow=0,f;
        for(int& i=cur[x];i<G[x].size();i++){
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return flow;
    }
    int Maxflow(int s,int t){
        this->s=s;
        this->t=t;
        int flow=0;
        while(BFS()){
            memset(cur,0,sizeof cur);
            flow+=DFS(s,INF);
        }
        return flow;
    }
}ac;
struct Node{
    int cost,time;
}nod[300];

int shop[300][300];
int mon[300];
int TT,n,m,T,S,L;

int judge(int n,int m,int lim){
    T=n+m+1;
    S=0;
    int tot=0;
    ac.init(m+n+2,S,T);
    for(int i=1;i<=m;i++){
        for(int j=1;j<=shop[i][0];j++)
            ac.addEdge(i,m+shop[i][j],INF);
        ac.addEdge(S,i,mon[i]);
        tot+=mon[i];
    }
    for(int i=1;i<=n;i++){
        if(nod[i].time<=lim){
            ac.addEdge(i+m,T,nod[i].cost);
        }
        else ac.addEdge(i+m,T,INF);
    }
    tot-=ac.Maxflow(S,T);
    return tot;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);

    cin>>TT;
    for(int t=1;t<=TT;t++){
        scanf("%d%d%d",&n,&m,&L);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&nod[i].cost,&nod[i].time);
        }
        for(int i=1;i<=m;i++){
            scanf("%d %d",&mon[i],&shop[i][0]);
            for(int j=1;j<=shop[i][0];j++){
                scanf("%d",&shop[i][j]);
            }
        }
        int l=1,r=1000000000;
        int ans=0;
        while(l<r){
            int mid=(r+l)/2;
            ans=judge(n,m,mid);
            if(ans>=L){
                r=mid;
            }
            else l=mid+1;
        }
        ans=judge(n,m,l);//注意这里
        printf("Case #%d: ",t);
        if(ans>=L)
            printf("%d %d\n",l,ans);
        else
            puts("impossible");
    }
    return 0;
}