UVA - 10779 Collectors Problem(最大流)

题目大意：有N个人在收集贴纸。现在给出每个人所拥有的贴纸

然后1这个想要得到更多种类的贴纸，所以他要拿他的贴纸去跟别人换，换的条件是1张交换1张，且你所交换的那张贴纸的种类对方没有，你想要得到的贴纸的种类对方至少有2张，问最后这个人能得到多少种贴纸

解题思路：一个超级源点，连接贴纸的种类，容量为1这个人所拥有的该种类的贴纸数量

将所有贴纸的种类连接到超级汇点，容量为1

在弄出N-1个点，代表另外的人，如果该贴纸的种类对方没有，那么连边，容量为1，因为只需要1张就够了(贴纸 –>人)

如果该贴纸对方有至少两张，那么连边，容量为这个人的该贴纸的数量-1（至少要保留一张，人–>贴纸）

然后跑最大流

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge{
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow) {}
};

struct Dinic{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G
;
    bool vis
;
    int d
, cur
;

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
        }
        edges.clear();
    }

    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 BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        vis[s] = 1;
        d[s] = 0;

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = true;
                    d[e.to] = d[u] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a) {
        if (x == t || a == 0)
            return a;

        int flow = 0, f;
        for (int i = cur[x]; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[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 s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS()) {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }
};

Dinic dinic;
int n, m, source, sink, cas = 1;
int num[15][30];

void init() {
    scanf("%d%d", &n, &m);
    memset(num, 0, sizeof(num));
    source = 0; sink = n + m;

    int x, y;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &x);
        for (int j = 1; j <= x; j++) {
            scanf("%d", &y);
            num[i][y]++;
        }
    }

    dinic.init(sink);
    for (int j = 1; j <= m; j++) {
        dinic.AddEdge(source, j, num[1][j]);
        dinic.AddEdge(j, sink, 1);
    }

    for (int i = 2; i <= n; i++) 
        for (int j = 1; j <= m; j++) {
            if (num[i][j] > 1) dinic.AddEdge(m + i - 1, j, num[i][j] - 1);
            if (!num[i][j]) dinic.AddEdge(j, m + i - 1, 1);
        }

    int ans = dinic.Maxflow(source, sink);
    printf("Case #%d: %d\n", cas++, ans);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
    }
    return 0;
}