CROC 2016 - Final Round [Private, For Onsite Finalists Only] C. Binary Table FWT

C. Binary Table

题目连接：

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

Description

You are given a table consisting of n rows and m columns. Each cell of the table contains either 0 or 1. In one move, you are allowed to pick any row or any column and invert all values, that is, replace 0 by 1 and vice versa.

What is the minimum number of cells with value 1 you can get after applying some number of operations?

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 20, 1 ≤ m ≤ 100 000) — the number of rows and the number of columns, respectively.

Then n lines follows with the descriptions of the rows. Each line has length m and contains only digits '0' and '1'.

Output

Output a single integer — the minimum possible number of ones you can get after applying some sequence of operations.

Sample Input

3 4
0110
1010
0111

Sample Output

2

Hint

题意

给你一个nm的01矩阵，然后每次操作：你可以挑选任意的某一行或者某一列翻转，然后你需要使得整个矩阵的1的数量尽可能少，问你最少数量是多少。

题解：

首先2^nm这个算法很简单：暴力枚举横着怎么翻转，然后每一列O（1）判断就好了。

然后正解怎么做呢？

我们令ans[i]是异或i之后的1的个数是多少，那么ans[i] = sigma(cnt[i]*num[i^j)，cnt[i]表示列那个二进制为i的个数，num[i]表示二进制为i这个数的1的数量是多少。

这个很显然发现 i^(i^j) = i，这就是一个异或卷积的形式，用FWT加速计算就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = (1<<20)+6;
int n,m,cnt[maxn];
long long x1[maxn],x2[maxn],ans[maxn];
string s[maxn];
long long t[maxn];
void utfxor(long long a[], int n) {
    if(n == 1) return;
    int x = n >> 1;
    for(int i = 0; i < x; ++ i) {
        t[i] = (a[i] + a[i + x]) >> 1;
        t[i + x] = (a[i + x] - a[i]) >> 1;
    }
    memcpy(a, t, n * sizeof(long long));
    utfxor(a, x); utfxor(a + x, x);
}

long long tmp[maxn];

void tfxor(long long a[], int n) {
    if(n == 1) return;
    int x = n >> 1;
    tfxor(a, x); tfxor(a + x, x);
    for(int i = 0; i < x; ++ i) {
        tmp[i] = a[i] - a[i + x];
        tmp[i + x] = a[i] + a[i + x];
    }
    memcpy(a, tmp, n * sizeof(long long));
}

void solve(long long a[],long long b[],int n)
{
    tfxor(a,n);
    tfxor(b,n);
    for(int i=0;i<n;i++) a[i]=1LL*a[i]*b[i];
    utfxor(a,n);
}

int main()
{
    for(int i=0;i<maxn;i++){
        int tmp = i;
        while(tmp){
            if(tmp&1)cnt[i]++;
            tmp>>=1;
        }
    }
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        cin>>s[i];
    for(int i=0;i<m;i++){
        int tmp = 0;
        for(int j=0;j<n;j++){
            if(s[j][i]=='1')tmp+=1<<j;
        }
        x1[tmp]++;
    }
    for(int i=0;i<(1<<n);i++)
        x2[i]=min(cnt[i],n-cnt[i]);
    solve(x1,x2,1<<n);
    long long ans = 1e15;
    for(int i=0;i<(1<<n);i++)
        ans=min(ans,x1[i]);
    cout<<ans<<endl;
}