【后缀数组】【RMQ】HDU 6194

string string string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

 

Input

The first line contains an integer T (T≤100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.

∑length(s)≤2∗106

Output

For each test case, print the number of the important substrings in a line.

 

Sample Input

2

2

abcabc

3

abcabcabcabc

Sample Output

6 9

 Solution

题意：给一个串，问你其中正好出现k次的子串有多少个

思路：求出给的串的后缀数组和height数组，然后对于height数组中一个长度k-1的区间，如果两边的值比区间内的最小值（可以RMQ进行O(1)的查询）都要小，那这段区间就是对答案有贡献的。注意k=1时需要特殊处理一下，此时就是后缀数组派上用场的时候了。有不少细节在里面，具体请看代码。还有一个坑到哭泣的地方：由于是多组数据，一定要记得每次清空height数组（不清空的话如果这次的字符串比上一次的短，访问height[n+1]时预期的是0，但实际上是上次字符串的height值），否则要找十年才知道WA在哪。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 101111
using namespace std;
char s[MAXN];
int t1[MAXN],t2[MAXN],cc[MAXN],sa[MAXN],rnk[MAXN],height[MAXN];
int len;
bool cmp(int *y,int a,int b,int k)
{
    int a1=y[a];
    int b1=y[b];
    int a2=a+k>=len ? -1:y[a+k];
    int b2=b+k>=len ? -1:y[b+k];
    return a1==b1 && a2==b2;
}
int make_sa()
{
    int *x=t1,*y=t2;
    int m=30;
    for(int i=0; i<m; i++) cc[i]=0;
    for(int i=0; i<len; i++) ++cc[x[i]=s[i]-'a'];
    for(int i=1; i<m; i++) cc[i]+=cc[i-1];
    for(int i=len-1; i>=0; i--) sa[--cc[x[i]]]=i;

    for(int k=1; k<=len; k<<=1)
    {
        int p=0;
        for(int i=len-k; i<len; i++) y[p++]=i;
        for(int i=0; i<len; i++)
           if( sa[i]>=k ) y[p++]=sa[i]-k;

        for(int i=0; i<m; i++) cc[i]=0;
        for(int i=0; i<len; i++) ++cc[x[y[i]]];
        for(int i=1; i<m; i++) cc[i]+=cc[i-1];
        for(int i=len-1; i>=0; i--) sa[--cc[x[y[i]]]]=y[i];

        swap(x,y);
        m=1; x[sa[0]]=0;
        for(int i=1; i<len; i++)
          x[sa[i]]=cmp(y,sa[i],sa[i-1],k) ? m-1:m++;

        if( m>=len ) break;
    }
}
void make_height()
{
    for(int i=0; i<len; i++) rnk[sa[i]]=i;
    height[0]=0;
    int k=0;
    for(int i=0; i<len; i++)
    {
        if(!rnk[i]) continue;
        int j=sa[rnk[i]-1];
        if(k) k--;
        while(s[i+k]==s[j+k]) k++;
        height[rnk[i]]=k;
    }
}


int n,minl[100022][18];

void S_table(){
    int l = log(len) / log(2);
    for(int j = 1;j <= l;++j){
        for(int i = 1;i + (1 << (j - 1)) - 1 <= len;++i){
            minl[i][j] = min(minl[i][j-1],minl[i+(1<<(j-1))][j-1]);
        }
    }
}

int rmq(int l,int r){
    int k = log(r-l+1) / log(2);
    return min(minl[l][k],minl[r-(1<<k)+1][k]);
}


int T,k,cnt[30];

int main(){
    cin >> T;
    while(T--){
        int ans = 0;
        memset(height,0,sizeof height);    /**< !!! */
        scanf("%d",&k);
        scanf("%s",s); len = strlen(s);
        make_sa();
        make_height();
        n = len - 1;
        for(int i = 0;i < 100022;++i) minl[i][0] = 1e9;
        for(int i = 1;i <= len;++i)
            minl[i][0] = height[i];
        S_table();


        if(k == 1){
            for(int i = 1;i <= n+1;++i){
                int tmp = len - sa[i-1];
                if(height[i-1] < tmp && height[i] < tmp)
                    ans += tmp - max(height[i-1],height[i]);
            }
            printf("%d
",ans);
            continue;
        }

        for(int i = 1;i <= n - k + 2;++i){
            int tmp = rmq(i,i+k-2);
            if(height[i-1] < tmp && height[i+k-1] < tmp)
                ans += tmp - max(height[i-1],height[i+k-1]);
        }

        printf("%d
",ans);
    }
}