https://vjudge.net/problem/UVA-10779#author=0

網絡流

1.Bob向他有的貼紙連邊，流量為他有的貼紙數量

2.每一種貼紙向匯點連流量為1的邊

3.其余人，如果沒貼紙i，由i向這個人連一條流量為1的邊

4.如果貼紙i數量>1，由這個人向i連一條流量為數量-1的邊

#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;
const int N = 45;
const int M = 650;

#define
 
 gc getchar()
#define oo 999999999

int n, m, now, S, T, TI;
int head[N], now_head[N], calc[N], dis[N];
struct Node{
    int u, v, flow, nxt;
}G[M];
queue <int> Q;

inline int read(){
    int x = 0; char c = gc;
    while(c < ‘0‘ || c > ‘9‘) c = gc;
    while(c >= ‘0‘ && c <= ‘
 
9‘) x = x * 10 + c - ‘0‘, c = gc;
    return x;
}

inline void add(int u, int v, int flow){
    G[now].v = v; G[now].flow = flow; G[now].nxt = head[u]; head[u] = now ++;
}

inline bool bfs(){
    for(int i = S; i <= T; i ++) now_head[i] = head[i], dis[i] = -1;
    dis[S] = 0;
    while(!Q.empty()) Q.pop();
    Q.push(S);
    
 
while(!Q.empty()){
        int topp = Q.front();
        Q.pop();
        for(int i = head[topp]; ~ i; i = G[i].nxt){
            int v = G[i].v;
            if(dis[v] == -1 && G[i].flow > 0){
                dis[v] = dis[topp] + 1;
                if(v == T) return 1;
                Q.push(v);
            }        
        }
    }
    return 0;
}

int dfs(int now, int flow){
    if(now == T) return flow;
    int ret = 0;
    for(int & i = now_head[now]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(dis[v] == dis[now] + 1 && G[i].flow > 0){
            int f = dfs(v, min(G[i].flow, flow - ret));
            if(f) {G[i].flow -= f; G[i ^ 1].flow += f; ret += f; if(ret == flow) break;}
        }
    }
    if(ret != flow) dis[now] = -1;
    return ret;
}

inline int Dinic(){
    int ret = 0;
    while(bfs()) ret += dfs(S, oo);
    return ret;
}

int main()
{
    TI = read();
    for(int Ti = 1; Ti <= TI; Ti ++){
        n = read(); m = read(); now = 0;
        T = n + m + 1; S = 1;
        for(int i = 0; i <= T; i ++) head[i] = -1;
        for(int i = 1; i <= m; i ++) add(n + i, T, 1), add(T, n + i, 0);
        int k = read();
        for(int i = 1; i <= k; i ++){int im = read(); calc[im] ++;} 
        for(int i = 1; i <= m; i ++) if(calc[i]) add(1, n + i, calc[i]), add(n + i, 1, 0);
        for(int i = 1; i <= m; i ++) calc[i] = 0;
        for(int i = 2; i <= n; i ++){
            int k = read();
            for(int j = 1; j <= k; j ++){int im = read(); calc[im] ++;}
            for(int j = 1; j <= m; j ++)
                if(calc[j] > 1) add(i, n + j, calc[j] - 1), add(n + j, i, 0);
                else if(!calc[j]) add(n + j, i, 1), add(i, n + j, 0);
            for(int j = 1; j <= m; j ++) calc[j] = 0;
        }
        int answer = Dinic();
        printf("Case #%d: %d\n", Ti, answer);
    }


    return 0;
}
/*
2
2 5
6 1 1 1 1 1 1
3 1 2 2
3 5
4 1 2 1 1
3 2 2 2
5 1 3 4 4 3
*/

Collectors Problem