(二分图最大匹配)The Perfect Stall

题目链接：http://poj.org/problem?id=1274

入门题目，暖身。

改最大流模板的解法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 205;
typedef long long LL;
const int INF = 0xfffffff;
struct edge
{
	int to, cap, rev;
	edge(int to, int cap, int rev) :to(to), cap(cap), rev(rev){}
};

vector<edge> G[maxn*2];
bool used[maxn*2];
int N, K;
bool can[maxn][maxn];
void add_edge(int from, int to, int cap)
{
	G[from].push_back(edge(to, cap, G[to].size()));
	G[to].push_back(edge(from, 0, G[from].size() - 1));
}
int dfs(int v, int t, int f)
{
	if (v == t) return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); i++)
	{
		edge &e = G[v][i];
		if (!used[e.to] && e.cap>0)
		{
			int d = dfs(e.to, t, min(f, e.cap));
			if (d > 0)
			{
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
int max_flow(int s, int t)
{
	int flow = 0;
	for (;;)
	{
		memset(used, 0, sizeof(used));
		int f = dfs(s, t, INF);
		if (f == 0) return flow;
		flow += f;
	}
}

void solve()
{
	int s = N + K, t = s + 1;
	for (int i = 0; i < N; i++)	add_edge(s, i, 1);
	for (int i = 0; i < K; i++) add_edge(N + i, t, 1);
	for (int i = 0; i < N;i++)
	for (int j = 0; j < K; j++)
	{
		if (can[i][j]) add_edge(i, N + j, 1);
	}
	printf("%d\n", max_flow(s, t));
}
int main()
{
	//freopen("f:\\input.txt", "r", stdin);
	while (~scanf("%d%d",&N,&K))
	{
		memset(can, false, sizeof(can));
		for (int i = 0; i < N + K + 2; i++) G[i].clear();
		for (int i = 0; i < N; i++)
		{
			int num;
			scanf("%d", &num);
			for (int j = 0; j < num; j++)
			{
				int stall; scanf("%d", &stall);
				can[i][--stall] = true;
			}
		}
		solve();
	}
	return 0;
}

简单的实现方法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 505;
typedef long long LL;
const int INF = 0xfffffff;
int N, K;
int V;
vector<int > G[maxn];
int match[maxn];
bool used[maxn];
void add_edge(int u, int v)
{
	G[u].push_back(v);
	G[v].push_back(u);
}
bool dfs(int v)
{
	used[v] = true;
	for (int i = 0; i < G[v].size(); i++)
	{
		int u = G[v][i], w = match[u];
		if (w < 0 || !used[w] && dfs(w))
		{
			match[v] = u;
			match[u] = v;
			return true;
		}
	}
	return false;
}
int bipartite_matching()
{
	int res = 0;
	memset(match, -1, sizeof(match));
	for (int v = 0; v < V; v++)
	{
		if (match[v] < 0)
		{
			memset(used, 0, sizeof(used));
			if (dfs(v))
			{
				res++;
			}
		}
	}
	return res;
}
int main()
{
	//freopen("f:\\input.txt", "r", stdin);
	while (~scanf("%d%d", &N, &K))
	{
		V = N + K;
		for (int i = 0; i < V; i++) G[i].clear();
		for (int i = 0; i < N; i++)
		{
			int num;
			scanf("%d", &num);
			for (int j = 0; j < num; j++)
			{
				int stall; scanf("%d", &stall);
				add_edge(i, N+(--stall));
			}
		}
		int ans = bipartite_matching();
		printf("%d\n", ans);
	}
	return 0;
}