学步园

大意不在赘述。

思路：这是我有关图论的参考书上的一本书上的例题，我顺便把它写了一下。

具体的实现过程可以参考我的另一篇博客：http://blog.csdn.net/wall_f/article/details/8229629

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;


const int MAXN = 10010;
const int MAXM = 400010;
const int INF = 0x3f3f3f3f;


struct Edge
{
	int v, f;
	int next;
}edge[MAXM];

int n, nf, nd;
int cnt;

int s, t;

int first[MAXN], level[MAXN];
int q[MAXN];
char str[3000];

void init()
{
	cnt = 0;
	memset(first, -1, sizeof(first));
}

void read_graph(int u, int v, int f)
{
	edge[cnt].v = v, edge[cnt].f = f;
	edge[cnt].next = first[u], first[u] = cnt++;
	edge[cnt].v = u, edge[cnt].f = 0;
	edge[cnt].next = first[v], first[v] = cnt++;
}

int bfs(int s, int t)
{
	memset(level, 0, sizeof(level));
	level[s] = 1;
	int front = 0, rear = 1;
	q[front] = s;
	while(front < rear)
	{
		int x = q[front++];
		if(x == t) return 1;
		for(int e = first[x]; e != -1; e = edge[e].next)
		{
			int v = edge[e].v, f = edge[e].f;
			if(!level[v] && f)
			{
				level[v] = level[x] + 1;
				q[rear++] = v;
			}
		}
	}
	return 0;
}

int dfs(int u, int maxf, int t)
{
	if(u == t) return maxf;
	int ret = 0;
	for(int e = first[u]; e != -1; e = edge[e].next)
	{
		int v = edge[e].v, f = edge[e].f;
		if(level[v] == level[u] + 1 && f)
		{
			int Min = min(maxf-ret, f);
			f = dfs(v, Min, t);
			edge[e].f -= f;
			edge[e^1].f += f;
			ret += f;
			if(ret == maxf) return ret;
		}
	}
	return ret;
}

int Dinic(int s, int t)
{
	int ans = 0;
	while(bfs(s, t)) ans += dfs(s, INF, t);
	return ans;
}

void read_case()
{
	init();
	s = 0, t = nf+nd+n+n+1;
	for(int i = 1; i <= n; i++) //A,A1连一条边 
	{
		int A = nf+nd+i, A1 = nf+nd+n+i;
		read_graph(A, A1, 1);
	}
	for(int i = 1; i <= nf; i++) read_graph(s, i, 1); //源点与食物相连 
	for(int i = 1; i <= n; i++)
	{
		int f2, d2;
		int A = nf+nd+i, A1 = nf+nd+n+i; //拆点为A,A1 
		scanf("%d%d", &f2, &d2);
		while(f2--)
		{
			int v;
			scanf("%d", &v);
			read_graph(v, A, 1); //食物与A 
		}
		while(d2--)
		{
			int v;
			scanf("%d", &v);
			read_graph(A1, v+nf, 1); //A1与饮料 
		}
	}
	for(int i = 1; i <= nd; i++) read_graph(i+nf, t, 1); //汇点与饮料相连 
}

void solve()
{
	read_case();
	int ans = Dinic(s, t);
	printf("%d\n", ans);
}

int main()
{
	while(~scanf("%d%d%d", &n, &nf, &nd))
	{
		solve();
	}
	return 0;
}