CF662C

题解

行比较少，容易想到将每一列的状态压缩
在行操作固定的情况下，容易发现每一列的操作就是翻转\(0\)和\(1\)，要取最小方案，方案唯一
所以我们只需求出每一种操作的答案
如果操作的行的集合为\(S\)，那么对于状态为\(e\)的列，将会变成\(e \; xor \; S\)，同时产生\(e \; xor \; S\)的答案
如果\(s\)的答案记为\(b[s]\)，状态为\(s\)的列数量为\(a[s]\)
那么对于操作\(S\)，最后的答案为

而\(b[s]\)和\(a[s]\)数组都可以预处理出来

记\(N = 2^{20}\)
复杂度为\(O(NlogN)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2100005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
	return flag ? out : -out;
}
char S[22][100005];
int n,m;
LL A[maxn],B[maxn];
void fwt(LL* a,int n,int f){
	for (int i = 1; i < n; i <<= 1)
		for (int j = 0; j < n; j += (i << 1))
			for (int k = 0; k < i; k++){
				LL x = a[j + k],y = a[j + k + i];
				a[j + k] = x + y,a[j + k + i] = x - y;
				if (f == -1) a[j + k] /= 2,a[j + k + i] /= 2;
			}
}
int main(){
	n = read(); m = read();
	REP(i,n) scanf("%s",S[i] + 1);
	REP(j,m){
		int s = 0;
		REP(i,n) s = s << 1 | (S[i][j] - '0');
		A[s]++;
	}
	int maxv = (1 << n) - 1;
	for (int s = 0; s <= maxv; s++){
		int cnt = 0;
		for (int i = s; i; i >>= 1) cnt += (i & 1);
		B[s] = min(cnt,n - cnt);
	}
	int deg = 1;
	while (deg <= maxv) deg <<= 1;
	fwt(A,deg,1); fwt(B,deg,1);
	for (int i = 0; i < deg; i++) A[i] = A[i] * B[i];
	fwt(A,deg,-1);
	LL ans = INF;
	for (int i = 0; i <= maxv; i++) ans = min(ans,A[i]);
	printf("%lld\n",ans);
	return 0;
}