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 ( T100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer  k ( k1) which is described above;
the second line contain a string  s ( length(s)105).
It's guaranteed that  length(s)2106.
 

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
 

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define fuck(x) cout<<"["<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< 1) maxx =  max(maxx,H[i]);
            if(i+k-1 < n) maxx  = max(maxx,H[i+k]);
            if(minx > maxx) ans  += minx  - maxx;
//            cout<

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