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[GXOI/GZOI2019]与或和

作者:互联网

嘟嘟嘟


在GX大佬cyh的提议下,我就开了这道题。


看到位运算,就想到每一位单独考虑。
那么对于\(AND\)操作,我们只要找全是1的子矩阵个数。
对于\(OR\)操作,用子矩阵总数-全0子矩阵个数即可。
这样就有一个\(O(n ^ 4 logN)\)(\(N\)是值域)的做法了,可以拿到50分。
然后我就没想出来。


现在的瓶颈在于找全1的子矩阵。
观察发现,对于同一列的点,如果从下往上扫,以这个点为左上角的全1子矩阵的宽一定是单调不增的,于是对于每一列,我们维护一个单调栈,同时维护栈内元素的和即可。
复杂度\(O(n ^ 2logN)\)。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<vector>
#include<ctime>
#include<assert.h>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; (y = e[i].to) && ~i; i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e3 + 5;
const ll mod = 1e9 + 7;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) putchar('-'), x = -x;
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
	freopen("ha.in", "r", stdin);
	freopen("ha.out", "w", stdout);
#endif
}

int n, a[maxn][maxn];

In ll inc(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}

#define pr pair<int, int>
#define mp make_pair
#define fir first
#define sec second
pr st[maxn];
int b[maxn][maxn], rMax[maxn][maxn], top = 0;
In ll solve(int x, bool flg)
{
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j) b[i][j] = ((a[i][j] >> x) & 1) ^ flg;
	for(int i = 1; i <= n; ++i)
		for(int j = n; j; --j) rMax[i][j] = b[i][j] ? max(1, rMax[i][j + 1] + 1) : 0;
	ll ret = 0, tot = 0;
	for(int j = 1; j <= n; ++j)
	{
		tot = top = 0;
		for(int i = n; i; --i)
		{
			int tp = 1;
			while(top && st[top].fir >= rMax[i][j]) 
			{
				tot = inc(tot, mod - st[top].sec * st[top].fir % mod);
				tp += st[top].sec, --top;	
			}
			if(rMax[i][j])
			{
				tot = inc(tot, tp * rMax[i][j] % mod);
				st[++top] = mp(rMax[i][j], tp);
				ret = inc(ret, tot);
			}
		}
	}
	return ret;
}

int main()
{
//	MYFILE();
	n = read();
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j) a[i][j] = read();
	ll tot = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j) tot = inc(tot, i * j);
	ll ans1 = 0, ans2 = 0, tp = 1;
	for(int x = 0; x < 31; ++x, tp = (tp << 1) % mod)
	{
		ans1 = inc(ans1, solve(x, 0) * tp % mod);
		ans2 = inc(ans2, inc(tot, mod - solve(x, 1)) * tp % mod);
	}
	write(ans1), space, write(ans2), enter;
	return 0;
}

标签:int,ll,GZOI2019,maxn,GXOI,mod,include,define
来源: https://blog.51cto.com/u_15234622/2831549