POJ1087 A Plug for UNIX —— 最大流

题目链接：https://vjudge.net/problem/POJ-1087

A Plug for UNIX

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17861		Accepted: 6172

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：

有n个不同类型的插孔，m个需要充电的用电器，k种适配器，且每种适配器有无限个，适配器之间可以互相拼接（双向）。问：怎样安排，才能使得尽量多的用电器能充上电？

题解：

最大流问题。可知对于一个用电器，他要充电有两种方式，一种是直接将插头插到插孔上去，一种是通过适配器将插头与插孔相连。

1.建立超级源点，超级源点与每个用电器相连，且边的容量为1，表明这种用电器只有一台。

2.由于适配器之间可以互相拼接，所以对于每一对能够拼接的适配器，连上一条边，且这条边是双向的，容量为INF，因为题目说明了每种适配器都有无限个，所以这种搭配也有无限个。

3.用电器与插孔相连，以及用电器与适配器相连，且边的容量都为1，表示这种用电器只有一台。

4.适配器与插孔相连，且边的容量为INF，因为适配器有无限个。

5.建立超级汇点，且每个插孔与超级汇点相连，边的容量为1，表明只有一个插孔。

6.求最大流即可。

思考：

若每种适配器只有一个呢？

把每个适配器拆成两点，且内部连一条边，容量为1，使得流经这种适配器的流量限制在1之内，然后把INF都改成1。推广：如果限制了只有m个，那么边的容量就设置为m。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int mod = 1e9+;
 const int MAXN = 3e2+;

 int maze[MAXN][MAXN];
 int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
 int flow[MAXN][MAXN];

 int sap(int start, int end, int nodenum)
 {
     memset(cur, , sizeof(cur));
     memset(dis, , sizeof(dis));
     memset(gap, , sizeof(gap));
     memset(flow, , sizeof(flow));
     int u = pre[start] = start, maxflow = , aug = INF;
     gap[] = nodenum;

     while(dis[start]<nodenum)
     {
         loop:
         for(int v = cur[u]; v<nodenum; v++)
         if(maze[u][v]-flow[u][v]> && dis[u] == dis[v]+)
         {
             aug = min(aug, maze[u][v]-flow[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])
                 {
                     flow[u][v] += aug;
                     flow[v][u] -= aug;
                 }
                 aug = INF;
             }
             goto loop;
         }

         int mindis = nodenum-;
         for(int v = ; v<nodenum; v++)
             if(maze[u][v]-flow[u][v]> && mindis>dis[v])
             {
                 cur[u] = v;
                 mindis = dis[v];
             }
         if((--gap[dis[u]])==) break;
         gap[dis[u]=mindis+]++;
         u = pre[u];
     }
     return maxflow;
 }

 /*  建图模式：

                   —— —— —— —— —— ——
                  |                 |
     超级源点-->device-->adapter-->plug-->超级汇点
                          |  |
                           --
 */
 char plug[MAXN][], dev[MAXN][][], ada[MAXN][][];
 int main()
 {
     int n, m, k;
     scanf("%d", &n);
     for(int i = ; i<=n; i++) scanf("%s", plug[i]);
     scanf("%d", &m);
     for(int i = ; i<=m; i++) scanf("%s%s", dev[i][], dev[i][]);
     scanf("%d",&k);
     for(int i = ; i<=k; i++) scanf("%s%s", ada[i][], ada[i][]);

     memset(maze, , sizeof(maze));
     for(int i = ; i<=m; i++)   //dev-->plug
     for(int j = ; j<=n; j++)
     {
         if(!strcmp(dev[i][],plug[j])) maze[n+i][j] = ;
     }
     for(int i = ; i<=k; i++)   //ada-->ada
     for(int j = ; j<=k; j++)
     {
         if(i==j) continue;
         if(!strcmp(ada[i][],ada[j][])) maze[n+m+i][n+m+j] = INF;
         if(!strcmp(ada[j][],ada[i][])) maze[n+m+j][n+m+i] = INF;
     }
     for(int i = ; i<=m; i++)   //dev-->ada
     for(int j = ; j<=k; j++)
     {
         if(!strcmp(dev[i][],ada[j][])) maze[n+i][n+m+j] = ;
         if(!strcmp(dev[i][],ada[j][])) maze[n+i][n+m+j] = ;
     }
     for(int i = ; i<=k; i++)   //ada--plug
     for(int j = ; j<=n; j++)
     {
         if(!strcmp(ada[i][],plug[j])) maze[n+m+i][j] = INF;
         if(!strcmp(ada[i][],plug[j])) maze[n+m+i][j] = INF;
     }

     int start = , end = n+m+k+;
     for(int i = ; i<=m; i++) maze[start][n+i] = ; //超级源点-->dev
     for(int i = ; i<=n; i++) maze[i][end] = ;     //plug-->超级汇点

     cout<< m-sap(start, end, n+m+k+) <<endl;
 }