poj 3281 最大流

poj 3281

题意：每只奶牛喜欢吃若干种食物和饮料，问只吃一种食物和饮料的奶牛最多有多少只。

思路：将奶牛做点拆开，形成源点-食物-奶牛（左）-奶牛（右）-饮料-汇点的链，每条边的容量都是1，奶牛（左）-奶牛（右）限制了一只奶牛同时最多吃一种食物喝一种饮料。

然后跑最大流就可以了。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 1010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge{
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init(){
    tol = 0;
    memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0){
    edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];
    edge[tol].flow = 0;head[u] = tol++;
    edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];
    edge[tol].flow = 0;head[v]=tol++;
}
int sap(int start,int ed,int N){
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N){
        if(u == ed){
            int Min = INF;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                        Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to]){
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1;i = edge[i].next){
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag){
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1;i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}
int main() {
    int n, f, d;
    while(~scanf("%d %d %d", &n, &f, &d)) {
        init();
        for(int i = 1; i <= f; i++) addedge(1, n + n + i + 1, 1);
        for(int i = 1; i <= d; i++) addedge(i + 1 + n + n + f, 2 + n + n + f + d, 1);
        for(int i = 1; i <= n; i++) addedge(1 + i, 1 + n + i, 1);
        for(int i = 1; i <= n; i++){
            int fi, di;
            scanf("%d %d", &fi, &di);
            while(fi--){
                int a;
                scanf("%d", &a);
                addedge(n + n + 1 + a, 1 + i, 1);
            }
            while(di--){
                int a;
                scanf("%d", &a);
                addedge(1 + n + i, 1 + n + n + f + a, 1);
            }
        }
        printf("%d\n", sap(1, n + n + f + d + 2, n + n + f + d + 2));
    }
}