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 这个是一个比较裸的最大流的题目，建图也不是特别难，但是我就是被卡到了，虽然我更觉得是因为自己对于题解的依赖性太强了，这个不太好啊，所以给自己暂时定一个规矩，就是当天写的题目，要是不明白至少当天不能看题解。这个题目就是把所有的设备连到源点s，插座之间可以转化的之间相连，然后再把插座连到汇点。
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <string>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
struct edge
{
int u, v, c, f;
edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {}
};
vector<edge>e;
vector<int>G[maxn];
int level[maxn];//BFS分层，表示每个点的层数
int iter[maxn];//当前弧优化
int m, s, t;
void init(int n)
{
for (int i = 0; i <= n; i++)G[i].clear();
e.clear();
}
void add(int u, int v, int c)
{
e.push_back(edge(u, v, c, 0));
e.push_back(edge(v, u, 0, 0));
m = e.size();
G[u].push_back(m - 2);
G[v].push_back(m - 1);
}
void BFS(int s)//预处理出level数组
//直接BFS到每个点
{
memset(level, -1, sizeof(level));
queue<int>q;
level[s] = 0;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
for (int v = 0; v < G[u].size(); v++)
{
edge& now = e[G[u][v]];
if (now.c > now.f && level[now.v] < 0)
{
level[now.v] = level[u] + 1;
q.push(now.v);
}
}
}
}
int dfs(int u, int t, int f)//DFS寻找增广路
{
if (u == t)return f;//已经到达源点，返回流量f
for (int &v = iter[u]; v < G[u].size(); v++)
//这里用iter数组表示每个点目前的弧，这是为了防止在一次寻找增广路的时候，对一些边多次遍历
//在每次找增广路的时候，数组要清空
{
edge &now = e[G[u][v]];
if (now.c - now.f > 0 && level[u] < level[now.v])
//now.c - now.f > 0表示这条路还未满
//level[u] < level[now.v]表示这条路是最短路，一定到达下一层，这就是Dinic算法的思想
{
int d = dfs(now.v, t, min(f, now.c - now.f));
if (d > 0)
{
now.f += d;//正向边流量加d
e[G[u][v] ^ 1].f -= d;
//反向边减d，此处在存储边的时候两条反向边可以通过^操作直接找到
return d;
}
}
}
return 0;
}
int Maxflow(int s, int t)
{
int flow = 0;
for (;;)
{
BFS(s);
if (level[t] < 0)return flow;//残余网络中到达不了t，增广路不存在
memset(iter, 0, sizeof(iter));//清空当前弧数组
int f;//记录增广路的可增加的流量
while ((f = dfs(s, t, INF)) > 0)
{
flow += f;
}
}
return flow;
}
map<string, int>mp;
int main()
{
int n,id=2;
cin >> n;
s = 0, t = 1;
for(int i=1;i<=n;i++)
{
string ch;
cin >> ch;
mp[ch] = id++;
add(mp[ch], t, 1);
}
int m; cin >> m;
for(int i=1;i<=m;i++)
{
string cch, ch;
cin >> cch >> ch;
mp[cch] = id++;
if (mp[ch] == 0) mp[ch] = id++;
add(mp[cch], mp[ch], 1);
add(s, mp[cch], 1);
}
int k; cin >> k;
for(int i=1;i<=k;i++)
{
string cch, ch;
cin >> cch >> ch;
if (mp[cch] == 0) mp[cch] = id++;
if (mp[ch] == 0) mp[ch] = id++;
add(mp[cch], mp[ch], inf);
}
int ans = Maxflow(s, t);
printf("%d\n", m-ans);
return 0;
}
原文地址：https://www.cnblogs.com/EchoZQN/p/10797906.html