最大流-POJ-1087-A Plug for UNIX

A Plug for UNIX
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 15582 Accepted: 5305

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

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题意：
出题人为了凑个UNIX也是够拼的，题目大意是说，有n个插座，m种电器，k种转换器，其中每个插座都有适应的插头型号，每个电器也都有自己的插头型号，只有插头型号相符才能插上供电，同时，转换器可以将一种插头转换为另一种插头，现在问最少有多少个电器无法供上电。

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题解：
其实就是问最多有多少电器能够供上电，拿电器总数减一下就是答案了。
在网络流专题里所以直接想着怎么构图，我是用了一个map来记录string和编号的对应关系，每当出现一种新的插头型号就在map中写入其编号，然后最终图中是有源汇点和string点三种。接下来是统计插座信息，从源点到插座对应的string连一条上限为该string数量的边。然后是电器信息，我将这个放到了最后加入图，因为如果现在就加入，后面转换头中又出现了新的插头型号，汇点的编号就不好弄了。所以我先将转换头信息加入图中，对于str1,str2这样一个转换头，在图中从str2连一条无上限的边到str1。最后再从string点连一条上限为插头型号为该string的电器的数量的边。

---

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <string>
using namespace std;

const int MAXN=1100;
const int INF=0x7fffffff;
int n,m,k;
map<string,int>Book;
int num[MAXN],Rtotal,num1[MAXN];
int maze[MAXN][MAXN];
int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
int sap(int start,int end,int nodenum)
{
    memset(cur,0,sizeof(cur));
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    int u=pre[start]=start,maxflow=0,aug=-1;
    gap[0]=nodenum;
    while(dis[start]<nodenum)
    {
loop:
        for(int v=cur[u]; v<nodenum; v++)
            if(maze[u][v] && dis[u]==dis[v]+1)
            {
                if(aug==-1 || aug>maze[u][v])aug=maze[u][v];
                pre[v]=u;
                u=cur[u]=v;
                if(v==end)
                {
                    maxflow+=aug;
                    for(u=pre[u]; v!=start; v=u,u=pre[u])
                    {
                        maze[u][v]-=aug;
                        maze[v][u]+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        int mindis=nodenum-1;
        for(int v=0; v<nodenum; v++)
            if(maze[u][v]&&mindis>dis[v])
            {
                cur[u]=v;
                mindis=dis[v];
            }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}
int main(void)
{
    string str,str1;
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i=1; i<=n; i++)
    {
        cin >> str;
        if(Book[str]!=0)
        {
            num[Book[str]]++;
            continue;
        }
        else
        {
            Book[str]=++Rtotal;
            num[Rtotal]=1;
        }
    }
    for(int i=1;i<=Rtotal;i++)
        maze[0][i]=num[i];
    cin >> m;
    for(int i=1; i<=m; i++)
    {
        cin >> str >> str1;
        if(Book[str1]!=0)
        {
            num1[Book[str1]]++;
            continue;
        }
        else
        {
            Book[str1]=++Rtotal;
            num1[Rtotal]=1;
        }
    }
    cin >> k;
    for(int i=1;i<=k;i++)
    {
        cin >> str >> str1;
        if(Book[str]==0)
            Book[str]=++Rtotal;
        if(Book[str1]==0)
            Book[str1]=++Rtotal;
        maze[Book[str1]][Book[str]]=INF;
    }
    for(int i=1;i<=Rtotal;i++)
        maze[i][Rtotal+1]=num1[i];
    cout << m-sap(0,Rtotal+1,Rtotal+2) << endl;
    return 0;
}