bzoj 4443 [Scoi2015]小凸玩矩阵 二分答案+网络流

题面

题目传送门

解法

显然答案满足单调性，直接二分答案\(mid\)

考虑建一个类似于二分图的东西

因为每一行每一列只能被使用一次，所以拆点，容量为1

若\(a_{i,j}≤mid\)，那么在\(i\)和\(j\)相连，容量为1

限制整张网络容量为\(n-k+1\)

如果最大流为\(n-k+1\)，那么显然答案可行

其实不用拆点的，我是sb

代码

#include <bits/stdc++.h>
#define N 300
using namespace std;
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
    int next, num, c;
} e[N * N * 16];
int n, m, k, s, t, ss, cnt, l[N * 4], cur[N * 4], a[N][N];
void add(int x, int y, int c) {
    e[++cnt] = (Edge) {e[x].next, y, c};
    e[x].next = cnt;
}
void Add(int x, int y, int c) {
    add(x, y, c), add(y, x, 0);
}
bool bfs(int s) {
    for (int i = 1; i <= ss; i++) l[i] = -1;
    queue <int> q; q.push(s);
    while (!q.empty()) {
        int x = q.front(); q.pop();
        for (int p = e[x].next; p; p = e[p].next) {
            int k = e[p].num, c = e[p].c;
            if (c && l[k] == -1)
                l[k] = l[x] + 1, q.push(k);
        }
    }
    return l[t] != -1;
}
int dfs(int x, int lim) {
    if (x == t) return lim;
    int used = 0;
    for (int p = cur[x]; p; p = e[p].next) {
        int k = e[p].num, c = e[p].c;
        if (l[k] == l[x] + 1 && c) {
            int w = dfs(k, min(c, lim - used));
            e[p].c -= w, e[p ^ 1].c += w;
            if (e[p].c) cur[x] = p; used += w;
            if (used == lim) return lim;
        }
    }
    if (!used) l[x] = -1;
    return used;
}
int dinic() {
    int ret = 0;
    while (bfs(s)) {
        for (int i = s; i <= ss; i++)
            cur[i] = e[i].next;
        ret += dfs(s, INT_MAX);
    }
    return ret;
}
bool check(int mid) {
    s = 0, t = 2 * (n + m) + 1, ss = cnt = t + 1;
    if (cnt % 2 == 0) cnt++;
    for (int i = 0; i <= ss; i++) e[i].next = 0;
    Add(s, ss, n - k + 1);
    for (int i = 1; i <= n; i++)
        Add(i * 2 - 1, i * 2, 1), Add(ss, i * 2 - 1, 1);
    for (int i = 1; i <= m; i++)
        Add(2 * n + 2 * i - 1, 2 * n + 2 * i, 1), Add(2 * (n + i), t, 1);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (a[i][j] <= mid) Add(i * 2, 2 * n + 2 * j - 1, 1);
    int tx = dinic();
    return tx == n - k + 1;
}
int main() {
    read(n), read(m), read(k);
    int l = INT_MAX, r = -l, ans;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            read(a[i][j]), l = min(l, a[i][j]), r = max(r, a[i][j]);
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) ans = mid, r = mid - 1;
            else l = mid + 1;
    }
    cout << ans << "\n";
    return 0;
}