hdu 6194

string string string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 337    Accepted Submission(s): 75

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.

 

Output

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

 

Sample Input

22abcabc3abcabcabcabc

 

Sample Output

69

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

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后缀数组预处理出高度数组H，然后 用区间最小值-两端的最大值统计即可

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iomanip>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e5 + 20;const int mod = 1e9 + 7;const int INF = 0x3f3f3f3f;char s[MX];int SA[MX], R[MX], H[MX];int wa[MX], wb[MX], wv[MX], wc[MX];bool cmp(int *r, int a, int b, int l) {    return r[a] == r[b] && r[a + l] == r[b + l];}void Suffix(char *r, int m = 128) {    int n = strlen(r) +1;    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i++) wc[i] = 0;    for(i = 0; i < n; i++) wc[x[i] = r[i]]++;    for(i = 1; i < m; i++) wc[i] += wc[i - 1];    for(i = n - 1; i >= 0; i--) SA[--wc[x[i]]] = i;    for(j = 1, p = 1; p < n; j *= 2, m = p) {        for(p = 0, i = n - j; i < n; i++) y[p++] = i;        for(i = 0; i < n; i++) if(SA[i] >= j) y[p++] = SA[i] - j;        for(i = 0; i < n; i++) wv[i] = x[y[i]];        for(i = 0; i < m; i++) wc[i] = 0;        for(i = 0; i < n; i++) wc[wv[i]]++;        for(i = 1; i < m; i++) wc[i] += wc[i - 1];        for(i = n - 1; i >= 0; i--) SA[--wc[wv[i]]] = y[i];        for(t = x, x = y, y = t, p = 1, x[SA[0]] = 0, i = 1; i < n; i++) {            x[SA[i]] = cmp(y, SA[i - 1], SA[i], j) ? p - 1 : p++;        }    }    int k = 0; n--;    for(i = 0; i <= n; i++) R[SA[i]] = i;    for(i = 0; i < n; i++) {        if(k) k--;        j = SA[R[i] - 1];        while(r[i + k] == r[j + k]) k++;        H[R[i]] = k;    }}int mi[MX];int lowbit(int x){    return x&(-x);}void update(int x,  int n){    int lx,i;    while(x <=n){        mi[x]  = H[x];        lx  = lowbit(x);        for(int i = 1; i < lx; i<<=1)            mi[x] = min(mi[x],mi[x-i]);        x += lowbit(x);    }}int query(int x, int y){    int ans = INF;    while(y >= x){        ans  = min(H[y],ans);        y--;        for(;y-lowbit(y)>=x; y -= lowbit(y)){            ans= min(mi[y],ans);        }    }    return ans;}int main(){#ifdef __LOCAL__    freopen("input.txt","r",stdin);#endif // __LOCAL__    int T;    scanf("%d",&T);    while(T--){        int k;        scanf("%d",&k);        scanf("%s",s+1);        int n = strlen(s+1);        Suffix(s+1);        for(int i = 1; i <= n; i++)            update(i,n);//        cout<<s+1<<endl;//        for(int i = 1; i <= n; i++)//            cout<<H[i]<<" ";//        cout<<endl;        LL ans = 0;        for(int i = 1; i+k-1 <= n; i++){            int minx = i+1 <= i+k-1? query(i+1,i+k-1) : n-SA[i];            int maxx = 0;            if(i > 1) maxx =  max(maxx,H[i]);            if(i+k-1 < n) maxx  = max(maxx,H[i+k]);            if(minx > maxx) ans  += minx  - maxx;//            cout<<minx<<" "<<maxx<<endl;        }        cout<<ans<<endl;    }    return 0;}