紫书 例题11-7 UVa 753 （网络流最大流）

设一个源点， 到所有设备连一条弧， 容量为1， 然后设一个汇点， 所有插座到汇点连弧， 容量为1， 然后

转换器也连一条弧， 容量为1。 最后最大流就是答案。其中注意节点数要开大一些。

#include<cstdio>
#include<queue>
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 512; //注意总的节点数
struct Edge
{
	int from, to, cap, flow;
	Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow) {}
};
vector<Edge> edges;
vector<int> g[MAXN];
vector<string> name;
int h[MAXN], cur[MAXN], device[MAXN], target[MAXN];
int d[MAXN][MAXN], n, m, k, s, t;
int from[MAXN], to[MAXN]; 

int ID(string x)
{
	REP(i, 0, name.size())
		if(x == name[i])
			return i;
	name.push_back(x);
	return name.size() - 1;
}

void AddEdge(int from, int to, int cap)
{
	edges.push_back(Edge(from, to, cap, 0));
	edges.push_back(Edge(to, from, 0, 0));
	g[from].push_back(edges.size() - 2);
	g[to].push_back(edges.size() - 1);
}

bool bfs()
{
	memset(h, 0, sizeof(h));
	queue<int> q;
	q.push(s);
	h[s] = 1; //不要漏！ 

	while(!q.empty())
	{
		int u = q.front(); q.pop();
		REP(i, 0, g[u].size())
		{
			Edge& e = edges[g[u][i]];
			if(!h[e.to] && e.cap > e.flow)
			{
				h[e.to] = h[u] + 1;
				q.push(e.to);
			}
		}
	}

	return h[t];
}

int dfs(int x, int a)
{
	if(x == t || a == 0) return a;
	int flow = 0, f;
	for(int& i = cur[x]; i < g[x].size(); i++)
	{
		Edge& e = edges[g[x][i]];
		if(h[x] + 1 == h[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
		{
			e.flow += f;
			edges[g[x][i] ^ 1].flow -= f;
			flow += f;
			if((a -= f) == 0) break;
		}
	}
	return flow;
}

int solve()
{
	int ret = 0;
	while(bfs()) memset(cur, 0, sizeof(cur)), ret += dfs(s, 1e9);
	return ret;
}

int main()
{
	int T;
	scanf("%d", &T);

	while(T--)
	{
		memset(d, 0, sizeof(d));
		name.clear();
		string k1, k2;

		scanf("%d", &n);
		REP(i, 0, n)
		{
			cin >> k1;
			target[i] = ID(k1);
		}
		scanf("%d", &m);
		REP(i, 0, m)
		{
			cin >> k1 >> k2;
			device[i] = ID(k2);
		}
		scanf("%d", &k);
		REP(i, 0, k)
		{
			cin >> k1 >> k2;
			from[i] = ID(k1);
			to[i] = ID(k2);
		}

		int V = name.size();
		s = V, t = V + 1;

		edges.clear();
		REP(i, 0, V + 2) g[i].clear();
		REP(i, 0, m) AddEdge(s, device[i], 1);
		REP(i, 0, n) AddEdge(target[i], t, 1);
		REP(i, 0, k) AddEdge(from[i], to[i], 1e9);	

		printf("%d\n", m - solve());
		if(T) puts("");
	}

	return 0;
}