Codeforces Round #345 (Div. 2) E. Table Compression（并查集）

传送门

首先先从小到大排序，如果没有重复的元素，直接一个一个往上填即可，每一个数就等于当前行和列的最大值 + 1

如果某一行或列上有重复的元素，就用并查集把他们连起来，很（不）显然，处于同一行或列的相同元素始终应该保持一样的，然后再一个一个往上填

#include <bits/stdc++.h>
#define N 1000007
#define fi first
#define se second

using namespace std;

pair <int, pair<int, int> > A[N];
map <int, int> X, Y;

int n, m;
int Hx[N], Hy[N], ans[N], f[N];

inline int find(int x)
{
	return x == f[x] ? x : f[x] = find(f[x]);
}

inline void uni(int x, int y)
{
	x = find(x);
	y = find(y);
	if(x != y) f[x] = y;
}

int main()
{
	int i, j = -1, k, x, y, p;
	scanf("%d %d", &n, &m);
	for(i = 0; i < n * m; i++)
	{
		f[i] = i;
		scanf("%d", &A[i].first);
		A[i].se.fi = i / m;
		A[i].se.se = i % m;
	}
	sort(A, A + n * m);
	for(i = 0; i < n * m; i++)
	{
		if(i != n * m - 1 && A[i].fi == A[i + 1].fi) continue;
		for(k = j + 1; k <= i; k++)
		{
			x = A[k].se.fi;
			y = A[k].se.se;
			p = A[k].se.fi * m + A[k].se.se;
			Hx[x] = p;
			Hy[y] = p;
		}
		for(k = j + 1; k <= i; k++)
		{
			x = A[k].se.fi;
			y = A[k].se.se;
			p = A[k].se.fi * m + A[k].se.se;
			uni(Hx[x], p);
			uni(Hy[y], p);
		}
		for(k = j + 1; k <= i; k++)
		{
			x = A[k].se.fi;
			y = A[k].se.se;
			p = A[k].se.fi * m + A[k].se.se;
			p = find(p);
			ans[p] = max(ans[p], max(X[x], Y[y]) + 1);
		}
		for(k = j + 1; k <= i; k++)
		{
			x = A[k].se.fi;
			y = A[k].se.se;
			p = A[k].se.fi * m + A[k].se.se;
			p = find(p);
			X[x] = max(X[x], ans[p]);
			Y[y] = max(Y[y], ans[p]);
		}
		j = i;
	}
	for(i = 0; i < n * m; i++)
	{
		printf("%d ", ans[find(i)]);
		if(i % m == m - 1) puts("");
	}
	return 0;
}