UOJ #219 BZOJ 4650 luogu P1117 [NOI2016]优秀的拆分 (后缀数组、ST表)

连NOI Day1T1都不会做。。。看了题解都写不出来还要抄Claris的代码。。

题目链接: (luogu)https://www.luogu.org/problemnew/show/P1117

(bzoj)https://www.lydsy.com/JudgeOnline/problem.php?id=4650

(uoj)http://uoj.ac/problem/219

题解:

\(f[i]\)表示以\(i\)结束的\(AA\)型子串个数，\(g[i]\)表示以\(i\)开始的\(AA\)型子串个数

怎么求\(f,g\)?

打破思维定势，谁说必须要一个一个求呢

分长度来求

枚举长度\(L\), 处理所有长度为\(2L\)的\(AA\)型子串对\(f\)和\(g\)的贡献

如果每隔\(L\)的长度放一个打点计时器，呸，关键点

那么任何\(AA\)型子串都会经过两个相邻关键点

首先肯定要满足这两个关键位置上的字符一样

在这个基础上求出往前往后最多多少个一样的

这个就转化成了LCP和LCS问题，并且两个相邻关键点对\(f,g\)数组的影响是区间+1，使用差分前缀和解决

然后推一推就行了，注意+1-1不要推错

时间复杂度为调和级数，\(O(n\log n)\)

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
using namespace std;

const int N = 1<<15;
const int LGN = 15;
const int S = 26;
int log2[N+3];
struct SparseTable
{
	int n;
	int str[N+3];
	int rk[N+3];
	int tmp[N+3];
	int height[N+3];
	int h[N+3];
	int sa[N+3];
	int wb[N+3];
	int mini[N+3][LGN+3];
	void get_sa()
	{
		int *x = rk,*y = tmp;
		for(int i=0; i<=S; i++) wb[i] = 0;
		for(int i=1; i<=n; i++) wb[x[i]=str[i]]++;
		for(int i=1; i<=S; i++) wb[i] += wb[i-1];
		for(int i=n; i>=1; i--) sa[wb[x[i]]--] = i;
		int s = S,p = 0;
		for(int j=1; p<n; j<<=1)
		{
			p = 0;
			for(int i=n-j+1; i<=n; i++) y[++p] = i;
			for(int i=1; i<=n; i++) if(sa[i]>j) y[++p] = sa[i]-j;
			for(int i=1; i<=s; i++) wb[i] = 0;
			for(int i=1; i<=n; i++) wb[x[y[i]]]++;
			for(int i=1; i<=s; i++) wb[i] += wb[i-1];
			for(int i=n; i>=1; i--) sa[wb[x[y[i]]]--] = y[i];
			swap(x,y);
			p = 1; x[sa[1]] = 1;
			for(int i=2; i<=n; i++) x[sa[i]] = (y[sa[i]]==y[sa[i-1]] && y[sa[i]+j]==y[sa[i-1]+j]) ? p : ++p;
			s = p;
		}
		for(int i=1; i<=n; i++) rk[sa[i]] = i;
		for(int i=1; i<=n; i++)
		{
			h[i] = h[i-1]==0 ? 0 : h[i-1]-1;
			while(i+h[i]<=n && sa[rk[i-1]]+h[i]<=n && str[i+h[i]]==str[sa[rk[i]-1]+h[i]])
			{
				h[i]++;
			}
		}
		for(int i=1; i<=n; i++) height[i] = h[sa[i]];
		for(int i=1; i<=n; i++) mini[i][0] = height[i];
		for(int j=1; j<=LGN; j++)
		{
			for(int i=1; i+(1<<j)-1<=n; i++)
			{
				mini[i][j] = min(mini[i][j-1],mini[i+(1<<j-1)][j-1]);
			}
		}
	}
	int querymin(int lb,int rb)
	{
		int g = log2[rb-lb+1];
		return min(mini[lb][g],mini[rb-(1<<g)+1][g]);
	}
	int LCP(int x,int y)
	{
		if(x==y) return n-x+1;
		if(rk[x]>rk[y]) swap(x,y);
		return querymin(rk[x]+1,rk[y]);
	}
	void clear()
	{
		for(int i=1; i<=n; i++) str[i] = rk[i] = tmp[i] = height[i] = h[i] = sa[i] = wb[i] = 0;
		for(int i=1; i<=n; i++)
		{
			for(int j=0; j<=LGN; j++)
			{
				mini[i][j] = 0;
			}
		}
	}
} s1,s2;
llong f[N+3];
llong g[N+3];
char a[N+3];
int n;

int LCP(int x,int y) {return s1.LCP(x,y);}
int LCS(int x,int y) {return s2.LCP(n+1-x,n+1-y);}

void preprocess()
{
	log2[1] = 0; for(int i=2; i<=N; i++) log2[i] = log2[i>>1]+1;
}

void clear()
{
	s1.clear(); s2.clear();
	for(int i=0; i<=n+1; i++) a[i] = 0,f[i] = g[i] = 0ll;
}

int main()
{
	preprocess();
	int T; scanf("%d",&T);
	while(T--)
	{
		scanf("%s",a+1); n = strlen(a+1); for(int i=1; i<=n; i++) a[i]-=96;
		for(int i=1; i<=n; i++) s1.str[i] = a[i]; s1.n = n;
		s1.get_sa();
		for(int i=1; i<=n; i++) s2.str[i] = a[n+1-i]; s2.n = n;
		s2.get_sa();
		for(int i=1; i+i<=n; i++)
		{
			for(int j=i+i; j<=n; j+=i)
			{
				if(a[j]==a[j-i])
				{
					int lb = j-LCS(j,j-i)+1,rb = j+LCP(j,j-i)-1;
					lb = max(lb+i-1,j); rb = min(rb,j+i-1);
					if(lb<=rb)
					{
						f[lb]++; f[rb+1]--;
						g[lb-i-i+1]++; g[rb+1-i-i+1]--;
					}
				}
			}
		}
		for(int i=1; i<=n; i++) f[i] += f[i-1],g[i] += g[i-1];
		llong ans = 0ll;
		for(int i=1; i<n; i++)
		{
			llong tmp = f[i]*g[i+1];
			ans += tmp;
		}
		printf("%lld\n",ans);
		clear();
	}
	return 0;
}