[SCOI2015]小凸玩矩阵

Description

给定一个 (n imes m) 的矩阵，从中选出 (n) 个数，使得任意两数不在同一行或同一列。问第 (k) 大数字的最小值是多少。

Solution

二分答案，将大于等于 (mid) 的数看作 (1)，将小于 (mid) 的数看作 (0)，然后 ( ext{check}) 一下最大匹配是否超过 (K) 即可。

于是满心欢喜地提交，听取 (WA) 声一片。你真以为就这么简单？cy

发现这样求出来的是第 (K) 大数字的最大值，因为每次二分需要尽量的减少为 (1) 的数，也就是说，我们的答案是尽量地往上“怼”的。

只要将第 (K) 大转化为第 (N - K + 1) 小即可。

Code

#include <bits/stdc++.h>
using namespace std;

const int _ = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int N, M, K, A[255][255], T[255][255];
int tot, head[_], to[_ << 1], nxt[_ << 1], edge[_ << 1];

void addEdge(int x, int y, int z) {
	to[++tot] = y, edge[tot] = z, nxt[tot] = head[x], head[x] = tot;
	to[++tot] = x, edge[tot] = 0, nxt[tot] = head[y], head[y] = tot;
}

namespace netFlow {
int S, T, d[_], cur[_];
bool bfs() {
	memset(d, 0, sizeof(d));
	copy(head, head + N + M + 2, cur);
	queue<int> q;
	q.push(S);
	d[S] = 1;
	while (q.size()) {
		int x = q.front();
		q.pop();
		for (int i = head[x]; i; i = nxt[i])
			if (edge[i] && !d[to[i]]) {
				d[to[i]] = d[x] + 1;
				q.push(to[i]);
				if (to[i] == T) return 1;
			}
	}
	return 0;
}
int dfs(int x, int flow) {
	if (x == T) return flow;
	int rest = flow, k;
	for (int &i = cur[x]; i; i = nxt[i])
		if (edge[i] && d[to[i]] == d[x] + 1) {
			k = dfs(to[i], min(rest, edge[i]));
			if (!k) d[to[i]] = 0;
			edge[i] -= k;
			edge[i ^ 1] += k;
			rest -= k;
		}
	return flow - rest;
}
int dinic(int s, int t) {
	S = s, T = t;
	int flow = 0, maxFlow = 0;
	while (bfs())
		while (flow = dfs(S, INF)) maxFlow += flow;
	return maxFlow;
}
}

bool check(int mid) {
	tot = 1;
	memset(head, 0, sizeof(head));
	for (int i = 1; i <= N; ++i)
		for (int j = 1; j <= M; ++j)
			if (A[i][j] <= mid) addEdge(i, j + N, 1);
	for (int i = 1; i <= N; ++i) addEdge(0, i, 1);
	for (int j = 1; j <= M; ++j) addEdge(j + N, N + M + 1, 1);
	return netFlow::dinic(0, N + M + 1) >= K;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("matrix.in", "r", stdin);
	freopen("matrix.out", "w", stdout);
#endif
	int l = 0, r = 0;
	cin >> N >> M >> K;
	for (int i = 1; i <= N; ++i)
		for (int j = 1; j <= M; ++j) {
			cin >> A[i][j];
			r = max(r, A[i][j]);
		}
	K = N - K + 1;
	while (l < r) {
		int mid = (l + r) >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}
	cout << l << endl;
	return 0;
}