题意
在一个 $n imes m$ 的矩阵 $A$ 中取出 $n$ 个数,任意两个数不能在同一行或同一列,求取出的第 $k$ 大数的最小值。
数据范围
$1 leq K leq N leq M leq 250$,$1 leq A_{i,j} leq 10^9$
分析
如果只是简单的取数,很容易想到把行和列分别放入两个点集做二分图
但是这里要求了第 $k$ 大数最小,所以我们可以二分枚举答案,判断该值是否成立
这也就相当于判断在小于等于该值的数中,能否取出合法的 $n-k+1$ 个数
但如果是在小于该值的数中取出 $n-k$ 个数,这样判断是否可行
实际上与枚举出的值相同的数不一定只有一个,这样做不能保证答案的最优性
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; #define ll long long #define inf 0x3f3f3f3f #define N 255 #define M (N * N << 1) int n, m, k, ans, tot, maxflow; int g[N][N], h[N * N], d[N << 1], cur[N << 1]; int to[M], cap[M], nxt[M], head[N << 1]; void add(int u, int v, int w) { to[++tot] = v; cap[tot] = w; nxt[tot] = head[u]; head[u] = tot; to[++tot] = u; cap[tot] = 0; nxt[tot] = head[v]; head[v] = tot; } bool bfs(int s, int t) { memset(d, 0, sizeof d); queue<int> q; d[s] = 1; q.push(s); while (!q.empty()) { int x = q.front(); q.pop(); for (int i = head[x]; i; i = nxt[i]) if (!d[to[i]] && cap[i]) { d[to[i]] = d[x] + 1; q.push(to[i]); } } return d[t]; } int dfs(int x, int t, int res) { if (x == t || !res) return res; int flow = 0, f; for (int &i = cur[x]; i; i = nxt[i]) if (d[to[i]] == d[x] + 1 && (f = dfs(to[i], t, min(cap[i], res)))) { flow += f; res -= f; cap[i] -= f; cap[i ^ 1] += f; if (!res) break; } return flow; } bool dinic(int now) { int s = 0, t = n + m + 1; tot = 1; maxflow = 0; memset(head, 0, sizeof head); for (int i = 1; i <= n; i++) add(s, i, 1); for (int i = 1; i <= m; i++) add(n + i, t, 1); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) if (g[i][j] <= now) add(i, n + j, 1); while (bfs(s, t)) { memcpy(cur, head, sizeof cur); maxflow += dfs(s, t, inf); } if (maxflow > n - k) return true; return false; } int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { scanf("%d", &g[i][j]); h[++h[0]] = g[i][j]; } sort(h + 1, h + n * m + 1); int l = 1, r = n * m; while (l <= r) { int mid = (l + r) >> 1; if (dinic(h[mid])) ans = h[mid], r = mid - 1; else l = mid + 1; } printf("%d ", ans); return 0; }