A Plug for UNIX

題目連結：https://vjudge.net/problem/POJ-1087

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

題意:

這個題意挺難理解的，一開始輸入的是插座，然後輸入的是插頭，最後輸入的是轉換器。

轉換器後面有兩個字串，也就是說能夠把插頭為第一個的轉化為第二個字串，這裡插頭有無限個。

要求最大能讓多少插頭連上插座。

題解：

首先考慮建圖的兩邊，源點連每個插頭且邊權為1，然後每個插座連上匯點且邊權為1。

然後每個插座可以使X->Y(舉例)，我們就可以想讓X連一條邊權為無窮大的邊到Y。

最後跑一個最大流就是了。注意下輸入時對字串的處理。

注意陣列要開到500左右，因為點的最多可能是400。

我這裡建圖是反過來建的，為了方便~

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#define INF 99999999
#define t 500
using namespace std;

const int N = 505;
int head[N],d[N];
int tot,n,m,k,cnt;
struct Edge{
    int v,next,c;
}e[N<<1];
char str[N][30],s[30],tmp[30];
void adde(int u,int v,int c){
    e[tot].v=v;e[tot].next=head[u];e[tot].c=c;head[u]=tot++;
    e[tot].v=u;e[tot].next=head[v];e[tot].c=0;head[v]=tot++;
}
bool bfs(int S,int T){
    memset(d,0,sizeof(d));d[S]=1;
    queue <int > q;q.push(S);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(!d[v] && e[i].c>0){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[T]!=0;
}
int dfs(int s,int a){
    int flow=0,f;
    if(s==t || a==0) return a;
    for(int i=head[s];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(d[v]!=d[s]+1) continue ;
        f=dfs(v,min(a,e[i].c));
        if(f){
            e[i].c-=f;
            e[i^1].c+=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[s]=-1;
    return flow;
}
int Dinic(){
    int max_flow = 0;
    while(bfs(0,t)){
        max_flow+=dfs(0,INF);
    }
    return max_flow;
}
int num=0;
int Search(char *s){
    if(num==0){
        strcpy(str[1],s);
        num++;
        return 1;
    }
    for(int i=1;i<=num;i++){
        if(strcmp(str[i],s)==0) return i;
    }
    num++;
    strcpy(str[num],s);
    return num;
}
int main(){
    scanf("%d",&n);
    memset(head,-1,sizeof(head));
    for(int i=1;i<=n;i++){
        scanf("%s",s);
        int a=Search(s);
        adde(0,a,1);
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++){
        scanf("%s%s",tmp,s);
        int a=Search(s);
        adde(a,t,1);
    }
    scanf("%d",&k);
    for(int i=1;i<=k;i++){
        scanf("%s%s",tmp,s);
        int a=Search(tmp),b=Search(s);
        adde(b,a,INF);
    }
    printf("%d",m-Dinic());
    return 0;
}