标签：出现 ace freopen binary etc 一个 open ons 次数
注意到n很小，显然的做法是枚举每行是否翻转，然后O(m)统计Σmin(popcount(ai),n-popcount(ai))即可。考虑将每列是否翻转写成一个二进制数，那么翻转相当于让该二进制数与每列异或。统计每列各种状态的出现次数，将其设为cnt[i]，将min(popcount(i),n-popcount(i))设为g[i]，可以发现若翻转状态为j，答案即为Σcnt[i]·g[i^j]。这就是一个异或卷积，FWT就行了。
#include<bits/stdc++.h> using namespace std; int getbit(){char c=getchar();while (c<‘0‘||c>‘9‘) c=getchar();return c^48;} int read() { int 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<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N (1<<20) #define M 100010 #define P 1000000007 #define inv2 500000004 int n,m,a[M],cnt[N],g[N],ans; void FWT(int *a,int n,int op) { for (int i=2;i<=n;i<<=1) for (int j=0;j<n;j+=i) for (int k=j;k<j+(i>>1);k++) { int x=a[k],y=a[k+(i>>1)]; a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P; if (op) a[k]=1ll*a[k]*inv2%P,a[k+(i>>1)]=1ll*a[k+(i>>1)]*inv2%P; } } int main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(),m=read(); for (int i=0;i<n;i++) for (int j=1;j<=m;j++) a[j]|=getbit()<<i; for (int i=1;i<=m;i++) cnt[a[i]]++; for (int i=1;i<(1<<n);i++) g[i]=g[i^(i&-i)]+1; for (int i=1;i<(1<<n);i++) g[i]=min(g[i],n-g[i]); FWT(g,(1<<n),0),FWT(cnt,(1<<n),0); for (int i=0;i<(1<<n);i++) g[i]=1ll*g[i]*cnt[i]%P; FWT(g,(1<<n),1); ans=n*m; for (int i=0;i<(1<<n);i++) ans=min(ans,g[i]); cout<<ans; return 0; }
标签：出现 ace freopen binary etc 一个 open ons 次数
原文地址：https://www.cnblogs.com/Gloid/p/10271801.html
