B - Dining POJ - 3281 （網絡流 - 最大流 - 拆點建圖）（多路增廣 + 當前弧優化）

(多路增廣 + 當前弧優化)[模板]

B - Dining POJ - 3281 （網絡流 - 最大流 - 建圖）

• 題目鏈接：https://vjudge.net/contest/68128#problem/B
• 題意：農場主有n頭牛，有一天他做了F種food，D種drink，每頭牛都有自己的喜好（food和drink的種類），每種food和drink都只能給一頭牛。求如何使滿足喜好的牛的數量最大。
• 思路：建圖，即爲最大流問題，直接用dinic解：
  

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <math.h>
#define pi acos(-1)
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
const int INF = 0x3f3f3f3f;
const LL ll_INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 50000 + 10;
const int mod = 1e9 + 7;


int n, f, d;

struct Edge
{
    int to, next, cap;
}es[maxn];


struct Dinic
{
    int s, t, maxf;
    int head[maxn], cnt, nowedge[maxn];
    int dis[maxn];
    void init()
    {
        cnt = 0;
        memset(head, -1, sizeof(head));
    }

    void addedge(int u, int v)
    {
        es[cnt].to = v;
        es[cnt].cap = 1;
        es[cnt].next = head[u];
        head[u] = cnt++;

        es[cnt].to = u;
        es[cnt].cap = 0;
        es[cnt].next = head[v];
        head[v] = cnt++;
    }

    bool BFS()
    {
        queue<int> que;
        memset(dis, -1, sizeof(dis));
        dis[s] = 0;
        que.push(s);
        int cur, v;
        while(!que.empty())
        {
            cur = que.front(); que.pop();
            for(int i=head[cur]; i!=-1; i=es[i].next){
                v = es[i].to;
                if(dis[v]==-1 && es[i].cap>0){
                    dis[v] = dis[cur]+1;
                    que.push(v);
                }
            }
        }
        if(dis[t] == -1) return false;
        else return true;
    }

    int DFS(int cur, int low)
    {
        int res=0, add=0;// 多路增廣
        if(cur == t) return low;
        for(int i=nowedge[cur]; i!=-1; i=es[i].next){
            nowedge[cur] = i;
            if(dis[es[i].to]==dis[cur]+1
            && es[i].cap>0
            && (add = DFS(es[i].to, min(low, es[i].cap))) ){
                es[i].cap -= add;
                es[i^1].cap += add;
                low -= add;
                res += add;
                if(low == 0) break;
            }
        }
        if(res == 0) dis[cur] = 0;
        return res;
    }


    int MAXFLOW()
    {
        maxf=0;
        int tmp=0;
        while(BFS()){
            for(int i=s; i<=t; i++){
                nowedge[i] = head[i];
            }
            if(tmp = DFS(s, INF))
                maxf += tmp;
        }
        return maxf;
    }
}DC;

int main()
{
    DC.init();
    scanf("%d%d%d", &n, &f, &d);
    DC.s = 0;
    DC.t = 2*n+f+d+1;
    for(int i=1; i<=n; i++){
        int numf, numd, fv, dv;
        scanf("%d%d", &numf, &numd);
        for(int j=1; j<=numf; j++){
            scanf("%d", &fv);
            //addedge(s, fv);
            DC.addedge(fv, f+i);
        }
        for(int j=1; j<=numd; j++){
            scanf("%d", &dv);
            //addedge(f+2*n+dv, t);
            DC.addedge(f+n+i, f+2*n+dv);
        }
        //addedge(f+i, f+n+i);
    }
    for(int i=1; i<=f; i++){
        DC.addedge(DC.s, i);
    }
    for(int i=1; i<=n; i++){
        DC.addedge(f+i, f+n+i);
    }
    for(int i=1; i<=d; i++){
        DC.addedge(f+2*n+i, DC.t);
    }

    printf("%d\n", DC.MAXFLOW());
}