CodeForces - 662C Binary Table (FWT)

题意:给一个N*M的0-1矩阵,能够进行若干次操做,每次操做将一行或一列的0和1反转,求最后能获得的最少的1的个数.
分析:本题可用FWT求解.
由于其0-1反转的特殊性且\(N\leq20\),将每一列j视做一个N位二进制数\(A[j]\),则一共有M个N位数,则能够统计出每一个二进制数i的个数\(num[i]\).将全部的行反转操做组合也视做一个N位二进制数\(S\).
那么如何将本题与FWT结合? 首先根据异或运算的结合律:\(S\oplus A[j]=B\),则\(S = A[j] \oplus B\),\(B\)确定也是一个N位二进制数.处理出每一个二进制数对应最少的1的个数(由于咱们能够将某一列也反转)\(B[j]\),则对于每一种行操做的组合\(S\),最后获得最少的1的个数即 \(cnt(S) = \sum_{i\oplus j = S}(num[i]*B[j])\).用FWT计算出每一个操做组合\(S\)的最少1个数,最后扫一遍全部操做,求其最小值便可.

#include<bits/stdc++.h>
using namespace std;
const int MAXN = (1<<20)+5;
typedef long long LL;

void FWT(LL a[] ,int n){
    for (int d = 1 ; d < n ; d <<= 1){
        for (int m = d << 1 ,i = 0;i < n ; i+=m){
            for (int j = 0 ; j < d ; j++){
                LL x = a[i+j],y = a[i+j+d];
                a[i+j] = (x+y),a[i+j+d] = (x-y);
            }
        }
    }
}

void UFWT(LL a[],int n){
    for (int d = 1 ; d < n ; d<<=1){
        for (int m = d <<1, i = 0; i < n; i+=m){
            for (int j = 0 ; j < d ; j++){
                LL x = a[i+j],y = a[i+j+d];
                a[i+j] = (x+y)/2; a[i+j+d] = (x-y)/2;
            }
        }
    }
}
void solve(LL a[],LL b[],int n){
    FWT(a,n);
    FWT(b,n);
    for (int i = 0 ; i<n ; i++)
        a[i]=a[i]*b[i];
    UFWT(a,n);
}

LL A[MAXN],B[MAXN],num[MAXN];
char str[MAXN];

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int N,M;
    scanf("%d %d",&N, &M);
    for(int i=0;i<N;++i){
        scanf("%s",str);
        for(int j=0;j<M;++j){
            A[j] |= ((str[j]-'0')<<i);
        }
    }
    for(int i=0;i<M;++i) num[A[i]]++;
    for(int i=0;i<(1<<N);++i){
        int cnt = __builtin_popcount(i);
        B[i] = min(cnt,N-cnt);
    }
    int all = 1<<N;
    solve(num,B,all);
    LL ans = N*M;
    for(int i=0;i<all;++i){
        ans = min(ans,num[i]);
    }
    printf("%lld\n",ans);
    return 0;
}