HDU6661 Acesrc and String Theory【SA】

Acesrc and String Theory

Problem Description

Acesrc is a famous string theorist at Nanjing University second to none. He insists that we should always say an important thing k times. He also believes that every string that can be obtained by concatenating k copies of some nonempty string is splendid. So, he always teaches newcomers, ``String theory problems are important! String theory problems are important! ... String theory problems are important!"

Today, he wants to examine whether the newcomers remember his instruction. He presents a string consisting of lower case letters and asks them the number of splendid substrings of the presented string. No one can solve this problem, and they will be scolded for hours. Can you help them solve this problem?

Note that equal splendid substrings occurred in different positions should be counted separately

给出一个字符串\(s\)，问\(s\)中有多少个子串是有\(k\)个相同串拼接而成的

有个暴力的做法，枚举循环节长度然后哈希，复杂度\(O(\frac{n^2}{k})\)

现在考虑\(O(n\log{n})\)的做法，同样是枚举循环节长度\(len\)，然后我们枚举循环节的起始位置，假设现在起始位置是\(pos\)，那么我们先找从\(pos\)开始的\(k-1\)个循环节，需要保证每个循环节\(lcp(pos,pos+i\cdot len)\ge len\)

现在有了\(k-1\)个循环节，位置从\(L(pos)\)，到\(R(pos+(k-1)\cdot len-1)\)，现在要找符合条件的\(k\)循环子串，我们只要知道\(L\)和\(R+1\)的最长公共前缀\(lcp\)和以\(R\)结尾和以\(L+1\)结尾的最长公共后缀\(lcs\)，就能知道这个\(k-1\)循环节对答案的贡献，我们显然可以构造一个字符串\(s_{l-pre}s_{l-pre+1}\cdots s_l s_{l+1}\cdots s_r s_{r+1}\cdots s_{r+suf}\)，其中\(pre+suf==len && pre\le lcs && suf\le lcp\)，所以当\(lcs+lcp\ge len\)的时候，对答案的贡献是\(lcs+lcp-len+1\)，注意左边界和右边界的处理情况，有点小细节，防止重复计算

由于枚举循环节，复杂度为调和级数\(O(\sum_{i=1}^{n}\frac{n}{i})=O(n\log n)\)

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 3e5+7;
using LL = int_fast64_t;
LL ret;
int n,K,rk[2][MAXN],sec[MAXN],c[MAXN],sa[2][MAXN],height[2][MAXN],ST[2][MAXN][20];
char s[MAXN],t[MAXN];
void SA(int m, char *s, int *rk, int *sa, int *height){
    int *RK = rk, *SEC = sec;
    for(int i = 0; i <= m; i++) c[i] = 0;
    for(int i = 1; i <= n; i++) c[RK[i]=s[i]]++;
    for(int i = 1; i <= m; i++) c[i] += c[i-1];
    for(int i = n; i >= 1; i--) sa[c[RK[i]]--] = i;
    for(int k = 1; k <= n; k <<= 1){
        int p = 0;
        for(int i = n - k + 1; i <= n; i++) SEC[++p] = i;
        for(int i = 1; i <= n; i++) if(sa[i]>k) SEC[++p] = sa[i]-k;
        for(int i = 0; i <= m; i++) c[i] = 0;
        for(int i = 1; i <= n; i++) c[RK[SEC[i]]]++;
        for(int i = 1; i <= m; i++) c[i] += c[i-1];
        for(int i = n; i >= 1; i--) sa[c[RK[SEC[i]]]--] = SEC[i];
        swap(RK,SEC);
        p = 0;
        RK[sa[1]] = ++p;
        for(int i = 2; i <= n; i++) RK[sa[i]] = SEC[sa[i]]==SEC[sa[i-1]] and SEC[sa[i]+k]==SEC[sa[i-1]+k] ? p : ++p;
        if(p==n) break;
        m = p;
    }
    int k = 0;
    for(int i = 1; i <= n; i++) rk[sa[i]] = i;
    for(int i = 1; i <= n; i++){
        if(rk[i]==1) continue;
        if(k) k--;
        int j = sa[rk[i]-1];
        while(i+k<=n and j+k<=n and s[i+k]==s[j+k]) k++;
        height[rk[i]] = k;
    }
}
void build_ST(){
    for(int i = 1; i <= n; i++){
        ST[0][i][0] = height[0][i];
        ST[1][i][0] = height[1][i];
    }
    for(int j = 1; (1<<j) <= n; j++){
        for(int i = 1; (i+(1<<j))-1 <= n; i++){
            ST[0][i][j] = min(ST[0][i][j-1],ST[0][i+(1<<(j-1))][j-1]);
            ST[1][i][j] = min(ST[1][i][j-1],ST[1][i+(1<<(j-1))][j-1]);
        }
    }
}
int qmin(int tg, int L, int R){
    int d = log2(R-L+1);
    return min(ST[tg][L][d],ST[tg][R-(1<<d)+1][d]);
}
int lcp(int tg, int l, int r){
    int rkl = rk[tg][l], rkr = rk[tg][r];
    if(rkl>rkr) swap(rkl,rkr);
    return qmin(tg,rkl+1,rkr);
}
void calc(int pos, int len){
    for(int i = 1; i < K - 1; i++) if(lcp(0,pos,pos+i*len)<len) return;
    int L = pos, R = L + (K-1)  * len - 1;
    int LCP = min(len,lcp(0,L,R+1));
    int LCS = min(len-1,lcp(1,n+1-R,n+1-L+1));
    if(LCP+LCS>=len) ret += LCP + LCS - len + 1;
}
void solve(){
    ret = 0;
    scanf("%d %s",&K,s+1);
    n = strlen(s+1);
    if(K==1){
        printf("%I64d\n",1ll*n*(n+1)/2);
        return;
    }
    for(int i = 1; i <= n; i++) t[i] = s[n+1-i];
    SA(128,s,rk[0],sa[0],height[0]);
    SA(128,t,rk[1],sa[1],height[1]);
    build_ST();
    for(int len = 1; len <= n; len++){
        for(int i = 1; i <= n; i += len){
            if(i+(K-1)*len-1>=n) break;
            calc(i,len);
        }
    }
    printf("%I64d\n",ret);
}
int main(){
    int T;
    for(scanf("%d",&T); T; T--) solve();
    return 0;
}