POJ - 1087 A Plug for UNIX （网络流最大流 + 建图）
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：房间中有 n 个插座，m 个电器需要充电，k个转换器（a, b）可将插座 b 转换为插座 a ，问最少有多少电器不能充电？
思路：
超级源点与 n 个插座之间连边，流量为1；转换前的插座与转换后的插座连边，流量为inf；插座与电器连边，流量为1；电器与超级汇点连边，流量为1。
对比上一题Dining（牛、菜、饮料）为什么这一题的插座没有拆点：虽然每种插座只有一个，但是由于转换器的作用，插座可以由其他类型的插座转换而成，所以同一类型的插座可能被用到多次，所以不需要拆点。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1005;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
int head[N], tot, n, m, k, s, t, nn, mm;
struct node {
int u, v, w;
};
struct Edge {
int from, to, next, cap, flow;
}edge[M];
void init() {
tot = 2;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w, int rw = 0) {
edge[tot].from = u;
edge[tot].to = v;
edge[tot].cap = w;
edge[tot].flow = 0;
edge[tot].next = head[u];
head[u] = tot++;
edge[tot].from = v;
edge[tot].to = u;
edge[tot].cap = rw;
edge[tot].flow = 0;
edge[tot].next = head[v];
head[v] = tot++;
}
int Q[N];
int dep[N], cur[N], sta[N]; ///数组cur记录点u之前循环到了哪一条边
bool bfs(int s, int t, int n) {
int fron = 0, tail = 0;
memset(dep, -1, sizeof(dep[0]) * (n + 1));
dep[s] = 0;
Q[tail++] = s;
while(fron < tail) {
int u = Q[fron++];
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(edge[i].cap > edge[i].flow && dep[v] == -1) {
dep[v] = dep[u] + 1;
if(v == t) return true;
Q[tail++] = v;
}
}
}
return false;
}
int dinic(int s, int t, int n) {
int maxflow = 0;
while(bfs(s, t, n)) {
for(int i = 0; i <= n; ++i) cur[i] = head[i];
int u = s, tail = 0;
while(cur[s] != -1) {
if(u == t) {
int tp = inf;
for(int i = tail - 1; i >= 0; --i)
tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);
maxflow += tp;
for(int i = tail - 1; i >= 0; --i) {
edge[sta[i]].flow += tp;
edge[sta[i] ^ 1].flow -= tp;
if(edge[sta[i]].cap - edge[sta[i]].flow == 0)
tail = i;
}
u = edge[sta[tail] ^ 1].to;
}
else if(cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) {
sta[tail++] = cur[u];
u = edge[cur[u]].to;
}
else {
while(u != s && cur[u] == -1)
u = edge[sta[--tail] ^ 1].to;
cur[u] = edge[cur[u]].next;
}
}
}
return maxflow;
}
map<string, int>mp1, mp2;
string u[N], v[N];
int main() {
init();
string a, b;
scanf("%d", &nn);
s = 0;
n = m = 0;
for(int i = 1; i <= nn; ++i) {
cin >> a;
if(!mp1[a]) mp1[a] = ++n;
addedge(s, n, 1); ///超级源点与插座
}
scanf("%d", &mm);
for(int i = 1; i <= mm; ++i) {
cin >> u[i] >> v[i];
mp2[u[i]] = ++m;
}
scanf("%d", &k);
for(int i = 1; i <= k; ++i) {
cin >> a >> b;
if(!mp1[a]) mp1[a] = ++n;
if(!mp1[b]) mp1[b] = ++n;
addedge(mp1[b], mp1[a], inf); ///插座与转换器 相当于插座与插座
}
t = n + m + 1;
for(int i = 1; i <= mm; ++i) {
if(!mp1[v[i]]) continue;
addedge(mp1[v[i]], mp2[u[i]] + n, 1); ///插座与电器
addedge(mp2[u[i]] + n, t, 1); ///电器与超级汇点
}
printf("%d\n", mm - dinic(s, t, t + 1));
mp1.clear(), mp2.clear();
return 0;
}
