2018icpc南京现场赛-I Magic Potion(最大流)

题意：

n个英雄，m个怪兽，第i个英雄可以打第i个集合里的怪兽，一个怪兽可以在多个集合里

有k瓶药水，每个英雄最多喝一次，可以多打一只怪兽，求最多打多少只

n,m,k<=500

思路：

最大流，建图方式：

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
//#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int head[maxn],d[maxn];//层
int ver[maxm],edge[maxm],Next[maxm];//edge[i]: c for edge_i
int n, m, s, t, tot, maxflow;
queue<int>q;
int vis[maxn];//出现过
void add(int x, int y, int z){
    ver[++tot]=y,edge[tot]=z,Next[tot]=head[x],head[x]=tot;

    ver[++tot]=x,edge[tot]=,Next[tot]=head[y],head[y]=tot;
}

bool bfs(){
    mem(d,);
    while(!q.empty())q.pop();
    q.push(s);
    d[s]=;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(int i = head[x]; i; i = Next[i]){
            if(edge[i] && !d[ver[i]]){
            q.push(ver[i]);
            d[ver[i]] = d[x] + ;
            if(ver[i] == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow){
    if(x==t) return flow;
    int rest = flow, k;
    for(int i = head[x]; i; i = Next[i]){
        if(edge[i] && d[ver[i]] == d[x]+){
            k = dinic(ver[i], min(rest, edge[i]));
            if(!k) d[ver[i]] = ;
            edge[i] -= k;
            edge[i^] += k;
            rest -= k;
        }
    }
    return flow - rest;
}

int main(){
    int num;
    scanf("%d %d %d", &n, &m, &num);
    tot = ;
    //scanf("%d %d", &s, &t);
    s = ;
    t = n+m+;
    add(s,,num);
    for(int i = ; i <= n; i++){
        int sz, x;
        add(s,+i,);
        add(,+i,);
        scanf("%d", &sz);
        for(int j = ; j <= sz; j++){
            scanf("%d", &x);
            add(+i,+n+x,);
        }
    }
    for(int i = ; i <= m; i++){
        add(+n+i,t,);
    }
    int flow = ;
    maxflow=;
    while(bfs()){
        while(){
            flow = dinic(s,inf);
            if(!flow)break;
            maxflow+=flow;
        }
    }
    printf("%d\n",maxflow);
    return ;
}
/*
3 5 0
4 1 2 3 5
2 2 5
2 1 2

5 10 2
2 3 10
5 1 3 4 6 10
5 3 4 6 8 9
3 1 9 10
5 1 3 6 7 10

 */