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
AC
建图：
- 将源点和电器需要插座建边，权值为需要的个数
- 将插座转换器建边，权值为INF（商店可以提供无数个）
- 将已有的插座和汇点相连，权值为该种类插座的数目
注意：
- 如果用邻接矩阵写，矩阵最小开到400，因为插座的种类最坏的情况是400种，已存在的插座100种，需要的插座100种，转换器100 + 100 = 200 种，所以最多400种
代码后面有极限数据
#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define N 300005
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;
int pre[405], a[405][405];
bool vis[405];
int inf = 0x3f3f3f3f;
int n, m, k;
// EK最大流
bool bfs(int s, int e) {
memset(pre, -1, sizeof(pre));
memset(vis, false, sizeof(vis));
vis[s] = true;
pre[s] = s;
queue<int> que;
que.push(s);
while (!que.empty()) {
int t = que.front();
que.pop();
for (int i = 0; i <= e; ++i) {
if (vis[i] || a[t][i] == 0) continue;
vis[i] = true;
pre[i] = t;
if (i == e) return true;
que.push(i);
}
}
return false;
}
ll solve(int s, int e) {
ll sum = 0;
while (bfs(s, e)) {
int MIN = inf;
for (int i = e; i != s; i = pre[i]) {
MIN = min(MIN, a[pre[i]][i]);
}
for (int i = e; i != s; i = pre[i]) {
a[pre[i]][i] -= MIN;
a[i][pre[i]] += MIN;
}
sum += MIN;
}
return sum;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
while (cin >> n) {
memset(a, 0, sizeof(a));
// 统计现有的插座种类和数目
map<string, int> mp1;
// 统计电器插头种类和数目
map<string, int> mp2;
// 对所有插座和插头进行标记
map<string, int> mp;
int now = 1;
// 读入插座
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
mp1[s]++;
if (mp[s] == 0) mp[s] = now++;
}
// 读入插头
cin >> m;
for (int i = 0; i < m; ++i) {
string s;
cin >> s >> s;
mp2[s]++;
if (mp[s] == 0) mp[s] = now++;
}
// 将源点和电器需要的插座相连，权值为需要的个数
map<string, int>::iterator it;
for (it = mp2.begin(); it != mp2.end(); ++it) {
a[0][mp[it->first]] = it->second;
}
// 读入转换器
cin >> k;
for (int i = 0; i < k; ++i) {
string s1, s2;
cin >> s1 >> s2;
if (mp[s1] == 0) mp[s1] = now++;
if (mp[s2] == 0) mp[s2] = now++;
// 因为商店可以提供无数个，权值设为 inf
a[mp[s1]][mp[s2]] = inf;
}
// 将现有的插座和汇点相连
for (it = mp1.begin(); it != mp1.end(); ++it) {
a[mp[it->first]][401] = it->second;
}
// 插座和插头最多400种，已有的100种，需要的100种，转换器100 + 100
ll ans = solve(0, 401);
printf("%d\n", m - ans);
}
return 0;
}- 数据
100
a0
a1
a2
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100
b0 c0
b1 c1
b2 c2
b3 c3
b4 c4
b5 c5
b6 c6
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b8 c8
b9 c9
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100
d0 e0
d1 e1
d2 e2
d3 e3
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