最大流问题

最小费用最大流：http://www.yhzq-blog.cc/最小费用最大流算法总结/

最大流：

http://poj.org/problem?id=3281

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<string.h>
#include<stdlib.h>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
int ln = 10, rn = 120, food = 230, drink = 340;
int n, m, f, d;
const int MAXN = 1100, MAXM = 1100000;

/*********最大流模板*********/
int head[MAXN], level[MAXN], cnt, s, t;
struct edgee{
    int to, cap, next;
}e[MAXM * 2];
//cnt从0开始，反向下标为（正向下标^1）
void add(int from, int to, int cap){
    e[cnt].to = to;        e[cnt].cap = cap;
    e[cnt].next = head[from];
    head[from] = cnt++;
    e[cnt].to = from;    e[cnt].cap = 0;
    e[cnt].next = head[to];
    head[to] = cnt++;
}
//bfs构建层次网络，从s起点，能到达t返回true
int bfs(){
    memset(level, 0, sizeof level);        level[s] = 1;
    queue<int>q;                        q.push(s);
    while (q.size()){
        int p = q.front(); q.pop();
        for (int i = head[p]; i != -1; i = e[i].next){
            if (level[e[i].to] == 0 && e[i].cap > 0){
                level[e[i].to] = level[p] + 1;    q.push(e[i].to);
                if (e[i].to == t)
                    return 1;
            }
        }
    }
    return 0;
}
//dfs沿着层次网络找增广路
int dfs(int u, int nf){
    if (u == t)return nf;
    int res = 0;
    for (int i = head[u]; ~i; i = e[i].next){
        if (level[e[i].to] != level[u] + 1 || !e[i].cap)continue;
        int d = dfs(e[i].to, min(nf, e[i].cap));
        res += d, nf -= d;
        e[i].cap -= d, e[i ^ 1].cap += d;
        if (nf == 0)return res;
    }
    level[u] = 0;
    return res;
}
//全局定义的s和t的最大流量
int maxflow(){
    int ans = 0, tf;
    while (bfs())
        ans += dfs(s, INF);
    return ans;
}

int main() {
    while (~scanf("%d %d %d", &n, &f, &d)){
        //初始化
        memset(head, -1, sizeof head);
        cnt = 0;
        s = 0, t = 500;
        //构图
        for (int i = 1; i <= f; i++)//食物与源点
            add(s, food + i, 1);
        for (int i = 1; i <= d; i++)//饮料与汇点
            add(drink + i, t, 1);
        for (int i = 1; i <= n; i++){
            add(ln + i, rn + i, 1);    //牛左与牛右
            scanf("%d %d", &f, &d);
            while (f--){            //食物与牛左
                scanf("%d", &m);
                add(food + m, ln + i, 1);
            }
            while (d--){            //饮料与牛右
                scanf("%d", &m);
                add(rn + i, drink + m, 1);
            }
        }
        printf("%d\n", maxflow());
    }
}