【UVALive - 3487】 Duopoly（网络流-最小割）

Description

The mobile network market in country XYZ used to be dominated by two large corporations, XYZ
Telecom and XYZ Mobile. The central government recently has realized that radio frequency spectrum
is a scarce resource and wants to regulate its usage. The spectrum currently in use is divided into
300,000 channels. Any wireless service provider who wishes to use certain spectrum should apply for
licenses on these channels. While some services may require use of multiple channels, a single channel
can not be shared by different services.
The central government wants to maximize its revenue from the spectrum by putting the channels
up to an auction. The only two bidders are XYZ Telecom and XYZ Mobile. They are allowed to place
bids on combinations of channels, through which their services can communicate with the customers.
Furthermore, the government stipulates that a company can only place at most one bid on a specific
channel.
The government can only accept a subset of the bids so none of them would conflict with each
other. However, officials soon find out that it is a difficult task to determine the winning bids in order
to maximize the revenue, and they are asking for your help.

Input
Standard input will contain multiple test cases. The first line of the input is a single integer T (1 ≤
T ≤ 10) which is the number of test cases. T test cases follow, each preceded by a single blank line.
Each test case has two bid description sections, which are for XYZ Telecom and XYZ Mobile,
respectively. Each section starts with an integer N (1 ≤ N ≤ 3, 000), which is the number of bids that
follow. The next N lines each contain the description for one bid, the first integer P (1 ≤ P ≤ 1, 000)
gives the price of that bid, followed by the channel numbers required by this service. A service would
require at least 1 channel and at most 32 channels. Each channel number is a positive integer and will
never exceed 300,000.

Output
Results should be directed to standard output. Start each case with ‘Case #:’ on a single line, where
# is the case number starting from 1. Two consecutive cases should be separated by a single blank
line. No blank line should be produced after the last test case.
For each test case, print the maximized revenue the government is able to collect by issuing licenses
on the channels.

Sample Input

2
3
45 1
51 2
62 3
4
54 1
15 2
33 3
2 4 5
5
20 1
18 2
23 4
54 3 5 6
17 7
4
36 1 2 3
28 5
47 4 7
16 6

Sample Output
Case 1:
169
Case 2:
139

 

 

【题意】

　　有两家公司都想向政府申请某些资源的使用权，并且他们都提供了一些申请列表，列表中含有申请费用和资源种类，同一家公司的申请列表之间不含有重复的资源。政府只可以完整地接受和拒绝谋一份申请列表，问政府的最大收益是多少。

 

【分析】

　　对于一组，新建一个点连源点（或汇点），流量为费用，再连向组内的全部点（流量为INF），最后跑一遍最小割即可。

 

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define INF 0xfffffff
#define Maxn 600010

int st,ed;
struct node
{
    int x,y,f,o,next;
}t[Maxn*2];int len;

int first[Maxn],dis[Maxn],sum;

int mymin(int x,int y) {return x<y?x:y;}

void ins(int x,int y,int f)
{
    t[++len].x=x;t[len].y=y;t[len].f=f;
    t[len].next=first[x];first[x]=len;t[len].o=len+1;
    t[++len].x=y;t[len].y=x;t[len].f=0;
    t[len].next=first[y];first[y]=len;t[len].o=len-1;
}

void init()
{
    len=0;
    memset(first,0,sizeof(first));
    st=300001;ed=st+1;sum=0;
    int x,now=ed;
    char cc;
    scanf("%d",&x);getchar();
    while(x--)
    {
        int c,y;
        scanf("%d",&c);getchar();
        ins(st,++now,c);sum+=c;
        while(scanf("%d%c",&y,&cc))
        {
            ins(now,y,INF);
            if(cc=='\n') break;
        }
    }
    scanf("%d",&x);getchar();
    while(x--)
    {
        int c,y;
        scanf("%d",&c);getchar();
        ins(++now,ed,c);sum+=c;
        while(scanf("%d%c",&y,&cc))
        {
            ins(y,now,INF);
            if(cc=='\n') break;
        }
    }
}

queue<int > q;
bool bfs()
{
    while(!q.empty()) q.pop();
    memset(dis,-1,sizeof(dis));
    q.push(st);dis[st]=0;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=first[x];i;i=t[i].next) if(t[i].f>0)
        {
            int y=t[i].y;
            if(dis[y]==-1)
            {
                dis[y]=dis[x]+1;
                q.push(y);
            }
        }
    }
    if(dis[ed]==-1) return 0;
    return 1;
}

int ffind(int x,int flow)
{
    if(x==ed) return flow;
    int now=0;
    for(int i=first[x];i;i=t[i].next) if(t[i].f>0)
    {
        int y=t[i].y;
        if(dis[y]==dis[x]+1)
        {
            int a=ffind(y,mymin(flow-now,t[i].f));
            now+=a;
            t[i].f-=a;
            t[t[i].o].f+=a;
        }
    }
    if(now==0) dis[x]=0;
    return now;
}

void max_flow()
{
    int ans=0;
    while(bfs())
    {
        ans+=ffind(st,INF);
    }
    printf("%d\n",sum-ans);
}

int main()
{
    int T,kase=0;
    scanf("%d",&T);
    while(T--)
    {
        init();
        printf("Case %d:\n",++kase);
        max_flow();
        if(T) printf("\n");
    }
    return 0;
}

 

　　最后一行输出空行报WA也是坑死了。

 

2016-05-27 13:22:21