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

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

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
",l,ans);
        else
            puts("impossible");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/pk28/p/5780811.html