cf622C(FWT)

题目链接

https://codeforces.com/problemset/problem/662/C

题意

给定 $n\times m$ 的 $01$ 矩阵,每次可以对行或列进行 $01$ 反转,问最少能使矩阵有多少个 $1$

$n\le 20,m\le 10^5$

题解

把每列当做一个数$a_i$ 的话,给每行反转相当于对每个数在该位上做一个异或,所以可以得知要求的是

把后面的 $min$ 看作一个序列,然后得到

直接 $FWT$ 之后取最值即可。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 1100005
#define nm 400005
using namespace std;
const double pi=acos(-1);
const ll inf=998244353;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}



int n,m,tot;
ll a[NM],b[NM],c[NM],ans;
char _s[NM];

void fwt(ll*a,int f=0){
for(int k=1;k<n;k<<=1)
for(int i=0;i<n;i+=k<<1)
for(int j=0;j<k;j++){
ll x=a[i+j],y=a[i+j+k];
a[i+j]=x+y;
a[i+j+k]=x-y;
}
if(f)inc(i,0,n-1)a[i]/=n;
}

int main(){
n=read();m=read();
inc(i,1,n){
scanf("%s",_s+1);
inc(j,1,m)if(_s[j]-'0')a[j]|=succ(i-1);
}
inc(i,1,m)c[a[i]]++;
tot=succ(n);
inc(i,0,tot-1)b[i]=__builtin_popcount(i),b[i]=min(b[i],n-b[i]);
n=tot;
fwt(c);fwt(b);
inc(i,0,n-1)b[i]=b[i]*c[i];
fwt(b,1);
ans=inf;
inc(i,0,n-1)ans=min(ans,b[i]);
printf("%lld\n",ans);
return 0;
}