POJ1087 A Plug for UNIX（网络流）

A Plug for UNIX

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【题意】有n个插座，m个电器，每个电器都对应一个插头，k个转换器，s1 s2表示插座s2可以转换成s1, 开始提供的插座只有一个插口，转换器有无数个，问最少有几个电器无法使用。【分析】这就是个最大流问题，关键在于建图。定义一个超级源点，超级汇点，源点指向n个插座，电器 指向汇点，插座指向对应的电器，容量均为1，s2指向s1,容量为无穷大，Dinic求最大流。还有就是 这个题目数据范围有问题，定105RE，后看讨论改的805.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=805;
const int M=300005;
int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
struct man
{
    int c,f;
}w[N][N];
int dis[N],n,m,k;
int s,ans,cnt;
int link;
map<string,int>p;
bool bfs()
{
    queue<int>q;
    memset(dis,0,sizeof(dis));
    q.push(s);
    dis[s]=1;
    while(!q.empty()){
        int v=q.front();q.pop();
        for(int i=0;i<=cnt;i++){
            if(!dis[i]&&w[v][i].c>w[v][i].f){
                q.push(i);
                dis[i]=dis[v]+1;
            }
        }
        if(v==1) return true;
    }
    return false;
}
int dfs(int cur,int cp)
{
    if(cur==1)return cp;
    int tmp=cp,tt;
    for(int i=0;i<=cnt;i++){
        if(dis[i]==dis[cur]+1 && tmp &&w[cur][i].c>w[cur][i].f){
            tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp));
            w[cur][i].f+=tt;
            w[i][cur].f-=tt;
            tmp-=tt;
        }
    }
    return cp-tmp;
}
void dinic()
{
    ans=0;
    while(bfs())ans+=dfs(s,inf);
}
int main(){

    string str,_str;
    while(~scanf("%d",&n)){
        memset(w,0,sizeof(w));
        p.clear();
        cnt=1;
        s=0;

        while(n--){
            cin>>str;
            p[str]=++cnt;
            w[0][p[str]].c=1;
        }
        cin>>m;
        int M=m;
        while(m--){
            cin>>str>>_str;
            p[str]=++cnt;
            if(!p[_str])p[_str]=++cnt;
            w[p[str]][1].c=1;
            w[p[_str]][p[str]].c=1;
        }
        cin>>k;
        while(k--){
            cin>>str>>_str;
            if(!p[str])p[str]=++cnt;
            if(!p[_str])p[_str]=++cnt;
            w[p[_str]][p[str]].c=inf;
        }

        dinic();
        cout<<M-ans<<endl;
    }
    return 0;
}