Sonya and Matrix Beauty CodeForces - 1080E (manacher)

Posted 2022-06-16 uid001

大意: 给定$nm$字符串矩阵. 若一个子矩形每一行重排后可以满足每行每列都是回文, 那么它为好矩形. 求所有好矩形个数.

 

一个矩形合法等价于每一行出现次数为奇数的最多只有一个字符, 并且对称的两行对应字符出现次数要完全相等.

那么直接暴力枚举左右边界, 把每个字符的出现次数$hash$一下, 这样就转化为给定序列, 求回文子串个数. 这是manacher算法经典应用, 套板子即可. 

暴力计算次数的话$O(26n^3)$竟然没卡过去, 改了好久最后位运算优化到$O(n^3)$才过.

#include <iostream>
#include <random>
#include <map>
#include <cstdio>
#include <algorithm>
#include <string.h>
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define hr puts("")
using namespace std;
typedef long long ll;
const int N = 1e3+10, P = 998244353;
 
int n, m, rad[N], fac[N];
int a[N], b[N], g[N];
char s[N][N];
 
void manacher(int *a, int n) 
	for (int i=1,j=0,k=-1; i<=n; i+=k) 
		while (a[i-j-1]==a[i+j+1]) ++j;
		rad[i] = j;
		for (k=1; k<=rad[i]&&rad[i-k]!=rad[i]-k; ++k) 
			rad[i+k] = min(rad[i-k], rad[i]-k);
		
		j = max(j-k, 0);
	

 
int calc(int *a, int n) 
	if (n<=0) return 0;
	b[1] = P+1;
	REP(i,1,n) 
		b[i*2] = a[i];
		b[i*2+1] = P+1;
	
	int ans = n;
	n = 2*n+1, b[n+1] = P+2;
	manacher(b,n);
	REP(i,1,n) ans += rad[i]/2;
	return ans;

 
int main() 
	fac[0] = 1;
	REP(i,1,30) fac[i] = fac[i-1]*991ll%P;
	scanf("%d%d", &n, &m);
	REP(i,1,n) scanf("%s",s[i]+1);
	ll ans = 0;
	REP(L,1,m)  
		REP(i,0,n) a[i] = g[i] = 0;
		REP(R,L,m)  
			int now = 0;
			REP(i,1,n) 
				a[i] = (a[i]+fac[s[i][R]-‘a‘])%P;
				g[i] ^= 1<<s[i][R]-‘a‘;
				if (g[i]&(g[i]-1)) 
					ans += calc(a+now,i-1-now);
					now = i;
				
			
			ans += calc(a+now,n-now);
		
	
	printf("%lld\n", ans);