K-th occurrence
作者：互联网
K-th occurrence
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 461 Accepted Submission(s): 125
For each query (l,r,k), please output the starting position of the k-th occurence of the substring SlSl+1...Sr in S.
Input The first line contains an integer T(1≤T≤20), denoting the number of test cases.
The first line of each test case contains two integer N(1≤N≤105),Q(1≤Q≤105), denoting the length of S and the number of queries.
The second line of each test case contains a string S(|S|=N) consisting of only lowercase english letters.
Then Q lines follow, each line contains three integer l,r(1≤l≤r≤N) and k(1≤k≤N), denoting a query.
There are at most 5 testcases which N is greater than 103.
Output For each query, output the starting position of the k-th occurence of the given substring.
If such position don't exists, output −1 instead.
Sample Input 2 12 6 aaabaabaaaab 3 3 4 2 3 2 7 8 3 3 4 2 1 4 2 8 12 1 1 1 a 1 1 1
Sample Output 5 2 -1 6 9 8 1
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = (1e5 + 10) * 100;
int M = 30;
int T, n, q, len, tot;
char s[maxn];
int sa[maxn], height[maxn];
int rk[maxn], tp[maxn];
int str[maxn], ton[maxn], root[maxn];
int f[300600][36];
struct tree {
int l, r, val;
} c[maxn];
inline void rsort() {
for (register int i = 0; i <= M; ++i)ton[i] = 0;
for (register int i = 1; i <= len; ++i)ton[rk[i]]++;
for (register int i = 1; i <= M; ++i)ton[i] = ton[i] + ton[i - 1];
for (register int i = len; i >= 1; --i)sa[ton[rk[tp[i]]]--] = tp[i];
}
inline void get_height() {
len++;
for (register int i = 1; i <= len; ++i)rk[i] = str[i], tp[i] = i;
rsort();
for (register int w = 1, p = 0; p < len && w <= len; M = p, w *= 2) {
p = 0;
for (register int i = len - w + 1; i <= len; ++i)tp[++p] = i;
for (register int i = 1; i <= len; ++i) {
if (sa[i] > w) {
tp[++p] = sa[i] - w;
}
}
rsort();
swap(rk, tp);
rk[sa[1]] = p = 1;
for (register int i = 2; i <= len; ++i) {
rk[sa[i]] = (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w]) ? p : ++p;
}
if (p >= len)break;
//printf("%d\n",p);
}
// for (register int i = 1; i <= n; ++i) {
// printf("sa[ %d ] = %d\n", i, sa[i]);
// }
int cur_len = len;
cur_len--;
int k = 0;
for (register int i = 1; i <= len; ++i)rk[sa[i]] = i;
// for(register int i=1;i<=len;++i){
// printf("rk[ %d ] = %d\n", i, rk[i]);
// }
for (register int i = 1; i <= cur_len; ++i) {
if (k)--k;
else k = 0;
int j = sa[rk[i] - 1];
while (str[i + k] == str[j + k])k++;
height[rk[i]] = k;
}
}
inline void init() {
for (register int i = 1; i <= len; ++i)f[i][0] = height[i];
for (register int j = 1; j <= 23; ++j) {
for (register int i = 1; i + (1 << j) - 1 <= len; ++i) {
f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
}
}
inline int query(int l, int r) {
int cur = log2(r - l + 1.0);
return min(f[l][cur], f[r - (1 << cur) + 1][cur]);
}
inline int update(int pre, int l, int r, int pos) {
int nx = ++tot;
c[nx] = c[pre];
c[nx].val++;
if (l == r)return nx;
int mid = l + r >> 1;
if (pos <= mid) {
c[nx].l = update(c[pre].l, l, mid, pos);
} else {
c[nx].r = update(c[pre].r, mid + 1, r, pos);
}
return nx;
}
inline int query(int pl, int pr, int l, int r, int val) {
if (l == r)return l;
int mid = l + r >> 1;
if (c[c[pr].l].val - c[c[pl].l].val >= val) {
return query(c[pl].l, c[pr].l, l, mid, val);
} else {
return query(c[pl].r, c[pr].r, mid + 1, r, val - (c[c[pr].l].val - c[c[pl].l].val));
}
}
int main() {
//freopen("1.txt", "r", stdin);
scanf("%d", &T);
while (T--) {
tot = 0;
scanf("%d%d", &n, &q);
scanf("%s", s + 1);
len = strlen(s + 1);
for (register int i = 0; i <= len + 1; ++i)root[i] = 0;
M = 30;
//printf("debug len = %d\n",len);
for (register int i = 1; i <= len; ++i)str[i] = s[i] - 'a' + 1;
str[len + 1] = 0;
get_height();
init();
for (register int i = 1; i <= len + 1; ++i) {
root[i] = update(root[i - 1], 1, n, sa[i]);
//printf("debug root[%d] = %d\n",i,root[i]);
}
int l, r, k;
while (q--) {
scanf("%d%d%d", &l, &r, &k);
int siz = r - l + 1;
//printf("debug rk[l] = %d\n",rk[l]);
l = rk[l];
int L = 1, R = l, ans_l, ans_r;
while (L <= R) {
int mid = L + R >> 1, cur;
cur = 100000;
if (mid + 1 <= l) cur = query(mid + 1, l);
if (cur >= siz) {
ans_l = mid;
R = mid - 1;
} else {
L = mid + 1;
}
// cur = query(mid, l);
// if (cur >= siz) {
// ans_l = mid;
// R = mid - 1;
// } else {
// L = mid + 1;
// }
}
//printf("debug ans_l = %d\n",ans_l);
L = l, R = len;
while (L <= R) {
int mid = L + R >> 1, cur;
cur = 100000;
if (l + 1 <= mid) cur = query(l + 1, mid);
if (cur >= siz) {
ans_r = mid;
L = mid + 1;
} else {
R = mid - 1;
}
// cur = query(l, mid);
// if (cur >= siz) {
// ans_l = mid;
// R = mid - 1;
// } else {
// L = mid + 1;
// }
}
// int pos = rk[l - 1];
// int L = 1, R = pos, LL = pos, RR = pos;
// while (L < R) {
// int mid = (L + R) / 2;
// if (query(mid, pos) >= r - l + 1)
// R = mid;
// else
// L = mid + 1;
// }
// if (query(L, pos) >= r - l + 1)
// LL = L - 1;
// if (pos < n - 1) {
// pos++;
// L = pos, R = n - 1;
// while (L < R) {
// int mid = (L + R) / 2;
// if (query(pos, mid) >= r - l + 1)
// L = mid + 1;
// else
// R = mid;
// }
// if (query(pos, L) < r - l + 1)
// L--;
// if (L >= pos && query(pos, L) >= r - l + 1)
// RR = L;
// }
//printf("debug ans_l = %d ans_r = %d\n",ans_l,ans_r);
if (ans_r - ans_l + 1 < k) {
puts("-1");
} else {
printf("%d\n", query(root[ans_l - 1], root[ans_r], 1, n, k));
}
}
}
return 0;
}
标签：occurrence,int,pos,mid,maxn,th,ans,query 来源： https://www.cnblogs.com/czy-power/p/11404138.html