A Plug for UNIX 分类： POJ 图论 函数 2015-08-10 14:18 2人阅读 评论(0) 收藏

A Plug for UNIX
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14786 Accepted: 4994

Description
You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucratic as possible.
Since the room was designed to accommodate reporters and journalists from around the world, it is equipped with electrical receptacles to suit the different shapes of plugs and voltages used by appliances in all of the countries that existed when the room was built. Unfortunately, the room was built many years ago when reporters used very few electric and electronic devices and is equipped with only one receptacle of each type. These days, like everyone else, reporters require many such devices to do their jobs: laptops, cell phones, tape recorders, pagers, coffee pots, microwave ovens, blow dryers, curling
irons, tooth brushes, etc. Naturally, many of these devices can operate on batteries, but since the meeting is likely to be long and tedious, you want to be able to plug in as many as you can.
Before the meeting begins, you gather up all the devices that the reporters would like to use, and attempt to set them up. You notice that some of the devices use plugs for which there is no receptacle. You wonder if these devices are from countries that didn’t exist when the room was built. For some receptacles, there are several devices that use the corresponding plug. For other receptacles, there are no devices that use the corresponding plug.
In order to try to solve the problem you visit a nearby parts supply store. The store sells adapters that allow one type of plug to be used in a different type of outlet. Moreover, adapters are allowed to be plugged into other adapters. The store does not have adapters for all possible combinations of plugs and receptacles, but there is essentially an unlimited supply of the ones they do have.

Input
The input will consist of one case. The first line contains a single positive integer n (1 <= n <= 100) indicating the number of receptacles in the room. The next n lines list the receptacle types found in the room. Each receptacle type consists of a string of at most 24 alphanumeric characters. The next line contains a single positive integer m (1 <= m <= 100) indicating the number of devices you would like to plug in. Each of the next m lines lists the name of a device followed by the type of plug it uses (which is identical to the type of receptacle it requires). A device name is a string of at most 24 alphanumeric
characters. No two devices will have exactly the same name. The plug type is separated from the device name by a space. The next line contains a single positive integer k (1 <= k <= 100) indicating the number of different varieties of adapters that are available. Each of the next k lines describes a variety of adapter, giving the type of receptacle provided by the adapter, followed by a space, followed by the type of plug.

Output
A line containing a single non-negative integer indicating the smallest number of devices that cannot be plugged in.

Sample Input

4
A
B
C
D
5
laptop B
phone C
pager B
clock B
comb X
3
B X
X A
X D

Sample Output

1

Source
East Central North America 1999
题意:给你插座,设备和适配器,求出最少有多少的设备无法插入
方法:最大流,建图,不过字符串最好要用map存,便于查找和赋值,注意认真读题,题意有点坑

#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int Max = 520;
int Map[520][520];
int n,m,k;
bool vis[520];
bool sign[520][520];
int t;
bool Dinic_BFS()//建立层次网络
{
    memset(vis,false,sizeof(vis));
    memset(sign,false,sizeof(sign));
    vis[0]=true;
    queue<int>Q;
    Q.push(0);
    while(!Q.empty()&&!vis[t])
    {
        int a=Q.front();
        Q.pop();
        for(int i=0; i<=t; i++)
        {
            if(!vis[i]&&Map[a][i])
            {
                vis[i]=true;
                sign[a][i]=true;
                Q.push(i);
            }
        }
    }
    if(vis[t])
    {
        return true;
    }
    return false;
}

int Dinic_DFS(int s,int sum)//增广
{
    if(s==t)
    {
        return sum;
    }
    int ant=sum;
    for(int i=0; i<=t; i++)
    {
        if(sign[s][i])
        {
            int ans=Dinic_DFS(i,min(sum,Map[s][i]));
            Map[s][i]-=ans;
            Map[i][s]+=ans;
            sum-=ans;
            if(!sum)
            {
                break;
            }
        }
    }
    return ant-sum;
}
int main()
{
    string s,c;
    int a=1,b=400;
    memset(Map,0,sizeof(Map));
    map<string,int>M;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        cin>>s;
        if(M.find(s)==M.end())
            M[s]=a++;
        Map[0][M[s]]=1;//插座与源点的容量为一
    }
    scanf("%d",&m);
    t=501;
    for(int i=0; i<m; i++)
    {
        cin>>s>>c;
        if(M.find(s)==M.end())
            M[s]=b++;
        if(M.find(c)==M.end())
        {
            M[c]=a++;
        }
        Map[M[c]][M[s]]=1;//在插座与设备之间的边的权值为一;
        Map[M[s]][t]=1;//设备与汇点的容量为一
    }
    scanf("%d",&k);
    for(int i=0; i<k; i++)
    {
        cin>>s>>c;
        if(M.find(s)==M.end())
        {
            M[s]=a++;
        }
        if(M.find(c)==M.end())
        {
            M[c]=a++;
        }
        Map[M[c]][M[s]]=INF;//插座与转换器之间的容量为INF,
    }
    int sum=0;
    while(Dinic_BFS())
    {
        sum+=Dinic_DFS(0,INF);
    }
    printf("%d\n",m-sum);
    return 0;
}

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