題干：

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

題目大意：

在一個會議室里有n種插座，每種插座一個,每個插座只能插一種以及一個電器（或者適配器）,有m個電器，每個電器有一個插頭需要插在相應一種插座上,不是所有電器都能在會議室找到相應插座,有k種適配器，每種適配器可以有無限多數量,每種適配器(a, b)可以把b類插座變為a類插座,問最后有多少個電器無法使用。

解題報告：

好像直接建圖并不難、、、讀清楚題就好了。

AC代碼：

#include<cstring> #include<cstdio> #include<algorithm> #include<iostream> #include<map> using namespace std; int tot; const int INF = 0x3f3f3f3f; struct Edge {int to,ne,w; } e[100005 * 2]; int head[10005]; int st,ed; int dis[10050],q[10005];//一共多少個點跑bfs，dis數組和q數組就開多大。 void add(int u,int v,int w) {e[++tot].to=v;e[tot].w=w;e[tot].ne=head[u];head[u]=tot; } bool bfs(int st,int ed) {memset(dis,-1,sizeof(dis));int front=0,tail=0;q[tail++]=st;dis[st]=0;while(front<tail) {int cur = q[front];if(cur == ed) return 1;front++;for(int i = head[cur]; i!=-1; i = e[i].ne) {if(e[i].w&&dis[e[i].to]<0) {q[tail++]=e[i].to;dis[e[i].to]=dis[cur]+1;}}}if(dis[ed]==-1) return 0;return 1; } int dfs(int cur,int limit) {//limit為源點到這個點的路徑上的最小邊權 if(limit==0||cur==ed) return limit;int w,flow=0;for(int i = head[cur]; i!=-1; i = e[i].ne) { if(e[i].w&&dis[e[i].to]==dis[cur]+1) {w=dfs(e[i].to,min(limit,e[i].w));e[i].w-=w;e[i^1].w+=w;flow+=w;limit-=w;if(limit==0) break;}}if(!flow) dis[cur]=-1;return flow; } int dinic() {int ans = 0;while(bfs(st,ed)) ans+=dfs(st,0x7fffffff);return ans; } map<string,int> mp,di; int id,n,m,k; int main() {char s1[55],s2[55];tot=1;memset(head,-1,sizeof head);cin>>n;for(int i = 1; i<=n; i++) {scanf("%s",s1);if(mp.find(s1) == mp.end()) mp[s1] = ++id;}int chatou = id;st = 0;cin>>m;for(int i = 1; i<=m; i++) {scanf("%s%s",s1,s2);if(di.find(s1) == di.end()) di[s1] = ++id;if(mp.find(s2) == mp.end()) mp[s2] = ++id;add(di[s1],mp[s2],1);add(mp[s2],di[s1],0);add(st,di[s1],1);add(di[s1],st,0);}cin>>k;for(int i = 1; i<=k; i++) {scanf("%s%s",s1,s2);add(mp[s1],mp[s2],INF);add(mp[s2],mp[s1],0);}ed = id+1;for(int i = 1; i<=chatou; i++) add(i,ed,1),add(ed,i,0);printf("%d\n",m - dinic());return 0; }

?

總結

以上是生活随笔為你收集整理的【POJ - 1087】A Plug for UNIX（建图，网络流最大流）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。