poj 3281 最大流 拆点构图，关键在于构图

/*

最大流，模版题目，要拆点

题目大意：给出N个牛喜欢的食物和饮料列表，每种食物和饮料只能用一种，求能满足多少只牛？

思路：把牛拆成两个点，连边e(i,n+i)=1

食物和i连i边e(food,i)=1,n+i和饮料连边e(n+i,drink)=1

食物和s连边e(s,i)=1, 饮料和t连边(i,）=1

*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

#define M 40010

int gap[M],dis[M],pre[M],cur[M];

int NE,NV;

int head[M];

struct Node{

int c,pos,next;

}E[999999];

#define FF(i,NV) for(int i=0;i<NV;i++)

int sap(int s,int t) {

//memset(pre,-1,sizeof(pre));

memset(dis,0,sizeof(int)*(NV+1));

memset(gap,0,sizeof(int)*(NV+1));

FF(i,NV) cur[i] = head[i];

int u = pre[s] = s,maxflow = 0,aug = -1;

gap[0] = NV;

while(dis[s] < NV) {

loop:for(int &i = cur[u]; i != -1; i = E[i].next) {

int v = E[i].pos;

if(E[i].c && dis[u] == dis[v] + 1) {

aug=min(aug,E[i].c);

pre[v] = u;

u = v;

if(v == t) {

maxflow += aug;

for(u = pre[u];v != s;v = u,u = pre[u]) {

E[cur[u]].c -= aug;

E[cur[u]^1].c += aug;

}

aug = 999999;

}

goto loop;

}

}

int mindis = NV;

for(int i = head[u]; i != -1 ; i = E[i].next) {

int v = E[i].pos;

if(E[i].c && mindis > dis[v]) {

cur[u] = i;

mindis = dis[v];

}

}

if( (--gap[dis[u]]) == 0)break;

gap[ dis[u] = mindis+1 ] ++;

u = pre[u];

}

return maxflow;

}

void addEdge(int u,int v,int c ) {

E[NE].c = c;E[NE].pos = v;

E[NE].next = head[u];head[u] = NE++;

E[NE].c = 0;E[NE].pos = u;

E[NE].next = head[v];head[v] = NE++;

}

int N, F, D,idx;

int main()

{

while (scanf("%d %d %d", &N, &F, &D) != EOF)

{

NV=F+2*N+D+2;

FF(i,NV) head[i] = -1;NE = 0;

memset(head, -1, sizeof(head));

idx = 0;

int a, b, f, d;

for (int i = 1; i <= N; ++i)

{

scanf("%d %d", &a, &b);

while (a--)

{

scanf("%d", &f);

addEdge(f, F + i, 1);

}

while (b--)

{

scanf("%d", &d);

addEdge(F + N + i, F + N + N + d, 1);

}

}

int source = 0, sink = N + N + F + D + 1;

//int n = sink + 1;

for (int i = 1; i <= F; ++i)

addEdge(source, i, 1);

for (int i = 1; i <= N; ++i)

addEdge(F + i, F + N + i, 1);

for (int i = 1; i <= D; ++i)

addEdge(F + N + N + i, sink, 1);

printf("%d\n", sap(source, sink));

}

return 0;

}