POJ 3281 最大流

一道隱含的最大流問題。
首先，把每種food當作一個節點，與頭部的s節點相連，每種drink當作一個節點，與結尾的t節點相連，這些路徑的容量都爲1（因爲每種food或者drink只能選一次）。然後如果在中間爲每頭牛建立一個節點，連接它可以用的food以及drink然後直接跑最大流的話，就會有一個問題，比如牛a可以喫foodA與foodB，喝drinkA和drinkB,那麼就會有兩個流經過這頭牛，那顯然是不對的，所以我們爲每頭牛建立兩個節點，一個連接food，一個連接drink，然後給這兩個節點中間連一條容量爲1的邊，確保一頭牛身上不會+2.

Dinic算法代碼

#include<iostream>
#include<math.h>
#include<iomanip>
#include <string>
#include <cstdio>
#include<stdio.h>
#include <cstring>
#include <algorithm>
#include <queue>
#include<vector>
#define INF 0x3f3f3f3f
#define N 500
#define ll long long
using namespace std;int stop;


#define drink(x) (2*n + f + x )
#define cowf(x) (f + x)
#define cowr(x) (f + n + x )
int s = 0,t;
vector<int>G[N];
struct Edge{
    int from,to,cap,flow;
    Edge(int a = 0,int b = 0,int c = 0,int d = 0){
        from = a,to = b,cap = c,flow = d;
    }
};
vector<Edge>edges;

void addedge(int from,int to,int cap){
    edges.push_back(Edge(from,to,cap,0));
    edges.push_back(Edge(to,from,0,0));
    int m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}

bool vis[N];
int deep[N],cur[N];

bool bfs(){
    memset(vis,0,sizeof(vis));
    queue<int>que;
    que.push(s);
    vis[s] = 1; deep[s] = 0;
    while (!que.empty())
    {
        int x = que.front();que.pop();
        for(int i = 0; i < G[x].size(); i++){
            Edge &e = edges[G[x][i]];
            if(!vis[e.to] && e.cap > e.flow){
                deep[e.to] = deep[x] + 1; 
                vis[e.to] = 1; 
                que.push(e.to); 
            }
        }
    }
    return vis[t];
}

int dfs(int x,int a){
    if(x == t || a == 0){
        return a;
    }
    int f,flow = 0;
    for(int& i = cur[x]; i<G[x].size(); i++){//遍歷每一條邊
        Edge &e = edges[G[x][i]];
        if(deep[x] + 1 == deep[e.to] && (f = dfs(e.to,min(a,e.cap - e.flow)) > 0)){
            e.flow += f;
            edges[G[x][i] ^ 1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0){
                break;
            }
        }
    }
    return flow;
} 

int MaxFlow(){
    int flow = 0;
    while (bfs())
    {
        memset(cur,0,sizeof(cur));
        flow += dfs(s,INF);
    }
    return flow;
}

int n,f,d;
int main(){
    int i,j,x,y,z;
    scanf("%d%d%d",&n,&f,&d);
    t = 2*n+f+d+1;
    for(i=1;i<=n;i++){
        addedge(cowf(i),cowr(i),1);
    }
    for(i=1;i<=f;i++){
        addedge(0,i,1);
    }
    for(i=1;i<=d;i++){
        addedge(drink(i),t,1);
    }
    for(i=1;i<=n;i++){
        scanf("%d%d",&x,&y);
        for(j=1;j<=x;j++){
            scanf("%d",&z);
            addedge(z,cowf(i),1);
        }

        for(j=1;j<=y;j++){
            scanf("%d",&z);
            addedge(cowr(i),drink(z),1);
        }
    }
    printf("%d\n",MaxFlow());
    return 0;
}