CF662C Binary Table

题目链接

solution

因为\(n\)比较小，所以我们可以\(2^n\)枚举每一行是不是翻转。然后对于每一列答案就唯一了。

对于每一列状态压缩，用\(B[i]\)表示\(i\)这个状态最小的\(1\)的个数(也就是这个状态里0和1更少的那个)。然后我们如果想把一列从状态\(x\)变成状态\(y\)，那么我们需要操作的行就是\(x\otimes y\)。用\(A[i]\)表示\(i\)这个状态的数量。用\(ans_x\)表示操作的行为\(x\)这个状态的时候的答案，就有\(ans_x=\sum\limits_{i\otimes j=x}A[x]B[y]\)。用\(FWT\)优化即可。

code

/*
* @Author: wxyww
* @Date:   2020-04-26 10:45:28
* @Last Modified time: 2020-04-26 11:21:07
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 1 << 20;
// #define int ll
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1; c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0'; c = getchar();
	}
	return x * f;
}
char s[N];
int n,m;

void FWT(ll *a,int xs) {
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < (1 << n);++j) {
			if((j >> i) & 1) continue;
			ll l = a[j],r = a[j | (1 << i)];
			a[j] = l + r;a[j | (1 << i)] = l - r;
		}
	}
	if(xs == -1) 
	for(int i = 0;i < (1 << n);++i)
		a[i] /= (1 << n);
}
ll a[N],cnt[N],A[N],B[N];
int main() {
	n = read(),m = read();
	for(int i = 1;i <= n;++i) {
		scanf("%s",s + 1);
		for(int j = 1;j <= m;++j)
			a[j] = (a[j] << 1) | (s[j] - '0');
	}

	for(int i = 1;i <= m;++i) A[a[i]]++;

	for(int i = 0;i < (1 << n);++i) {
		cnt[i] = (cnt[i >> 1]) + (i & 1);
		B[i] = min(cnt[i],n - cnt[i]);
	}

	FWT(B,1);FWT(A,1);
	for(int i = 0;i < (1 << n);++i) A[i] = A[i] * B[i];
	FWT(A,-1);

	ll ans = n * m;
	for(int i = 0;i < (1 << n);++i) {
		ans = min(ans,A[i]);
	}

	cout<<ans;
	return 0;
}