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「SCOI2015」小凸玩矩阵 - 二分图匹配

小方给小凸一个 NM N * M NM N \leq M )的矩阵 A A ,要求小秃从其中选出 N N 个数,其中任意两个数字不能在同一行或同一列,现小凸想知道选出来的 N N 个数中第 K K 大的数字的最小值是多少。

链接

BZOJ 4443

题解

二分第 K K 大的数是多少,然后建二分图,用行匹配列,检验匹配数是不是大于等于 NK+1 N - K + 1

注意,是第 K K 大,不是第 K K 小!

代码

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

const int MAXN = 250;
const int MAXM = 250;
const int MAXK = 250;
const int CACHE_FIX = 3;

struct Node;
struct Edge;

struct Node {
    Edge *firstEdge, *currentEdge;
    int level;
} nodes[MAXN + MAXM + 2];

struct Edge {
    Node *from, *to;
    int capacity, flow;
    Edge *next, *reversedEdge;

    Edge(Node *from, Node *to, int capacity) : from(from), to(to), capacity(capacity), flow(0), next(from->firstEdge) {}
};

struct Dinic {
    bool makeLevelGraph(Node *s, Node *t, int n) {
        for (int i = 0; i < n; i++) nodes[i].level = 0;

        std::queue<Node *> q;
        q.push(s);

        s->level = 1;

        while (!q.empty()) {
            Node *v = q.front();
            q.pop();

            for (Edge *e = v->firstEdge; e; e = e->next) {
                if (e->flow < e->capacity && e->to->level == 0) {
                    e->to->level = v->level + 1;
                    if (e->to == t) return true;
                    else q.push(e->to);
                }
            }
        }

        return false;
    }

    int findPath(Node *s, Node *t, int limit = INT_MAX) {
        if (s == t) return limit;

        for (Edge *&e = s->currentEdge; e; e = e->next) {
            if (e->flow < e->capacity && e->to->level == s->level + 1) {
                int flow = findPath(e->to, t, std::min(limit, e->capacity - e->flow));
                if (flow > 0) {
                    e->flow += flow;
                    e->reversedEdge->flow -= flow;
                    return flow;
                }
            }
        }

        return 0;
    }

    int operator()(int s, int t, int n) {
        int ans = 0;
        while (makeLevelGraph(&nodes[s], &nodes[t], n)) {
            for (int i = 0; i < n; i++) nodes[i].currentEdge = nodes[i].firstEdge;
            int flow;
            while ((flow = findPath(&nodes[s], &nodes[t])) > 0) ans += flow;
        }

        return ans;
    }
} dinic;

inline void addEdge(int from, int to, int capacity) {
    nodes[from].firstEdge = new Edge(&nodes[from], &nodes[to], capacity);
    nodes[to].firstEdge = new Edge(&nodes[to], &nodes[from], 0);

    nodes[from].firstEdge->reversedEdge = nodes[to].firstEdge, nodes[to].firstEdge->reversedEdge = nodes[from].firstEdge;
}

int n, m, k, a[MAXN + CACHE_FIX][MAXM], max;

void cleanUp() {
    for (int i = 0; i < n + m + 2; i++) {
        Edge *next;
        for (Edge *&e = nodes[i].firstEdge; e; next = e->next, delete e, e = next);
    }
}

bool check(int limit) {
    cleanUp();

    const int s = 0, t = n + m + 1;

    for (int i = 1; i <= n; i++) addEdge(s, i, 1);
    for (int j = 1; j <= m; j++) addEdge(n + j, t, 1);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (a[i - 1][j - 1] <= limit) addEdge(i, n + j, 1);
        }
    }

    int maxFlow = dinic(s, t, n + m + 2);
    // printf("%d\n", maxFlow);
    return maxFlow >= n - k + 1;
}

inline int solve() {
    static int tmp[MAXN * MAXM];
    for (int i = 0, c = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            tmp[c++] = a[i][j];
        }
    }
    std::sort(tmp, tmp + n * m);

    int *l = tmp, *r = std::unique(tmp, tmp + n * m) - 1;
    while (l < r) {
        int *const mid = l + ((r - l) >> 1);
        if (check(*mid)) r = mid;
        else l = mid + 1;
    }

    return *l;
}

int main() {
    // freopen("matrix.in", "r", stdin);
    // freopen("matrix.out", "w", stdout);

    scanf("%d %d %d", &n, &m, &k);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%d", &a[i][j]);
        }
    }

    printf("%d\n", solve());

    fclose(stdin);
    fclose(stdout);

    return 0;
}