The meaning of problems
Seeking interval l, r sub-string position in the original string k-th occurrence.
Links: https://vjudge.net/contest/322094#problem/C
Thinking
When the game with the suffix automaton wrote, TLE to end the game.
After learning suffix array and found this problem with the suffix array is also simpler to write.
We listed the sample aaabaabaaaab suffix Sort:
For example, we are l, r, k is 2,3,2, so first find 2 and 3 are represented as a string aa, height array suffix array represents the (sorted) adjacent to two of the longest common prefix of suffix length, think in this direction, [l, r] the substring must be a prefix of suffix. We first find RK [l] (l suffix starting), where rk [2] is 6, 6 may readily be found before and after the common prefix, if the common prefix> = (r-l + 1), these prefix has to meet the substring we're looking for. And you can find common prefix length suffix sorted is monotonic, the more closer here longer common prefix 6, away from the more common prefix 6 shorter half we can find upper and lower bounds, you can query table ST the height [rk [rk [y] x + 1] ~] minimum to find the longest common prefix length suffixes x and y suffix.
Find out after the next, these suffixes are to meet the conditions, then we advance all the sa [i] (i-th suffix sorted position in the original string) is inserted Chairman tree, and then find that a small k-bounds in can.
Code
#include<bits/stdc++.h>
const int N = 2e5 + 10;
using namespace std;
char s[N];
int len, M, rk[N], sa[N], tax[N], tp[N];
/*
sa[i]：排名为i的后缀的位置
rk[i]：从第i个位置开始的后缀的排名，把从第i个位置开始的后缀简称为后缀i
tp[i]：基数排序的第二关键字，意义与sa一样，即第二关键字排名为i的后缀的位置
tax[i]：i号元素出现了多少次。辅助基数排序
s：字符串，s[i]表示字符串中第i个字符串
*/
void jsort() //基数排序
{
for (int i = 0; i <= M; i++) tax[i] = 0;
for (int i = 1; i <= len; i++) tax[rk[i]]++;
for (int i = 1; i <= M; i++) tax[i] += tax[i - 1];
for (int i = len; i >= 1; i--) sa[ tax[rk[tp[i]]]-- ] = tp[i];
}
void suffixSort() //后缀排序
{
M = 75; //字符集的大小
for (int i = 1; i <= len; i++) rk[i] = s[i] - 'a' + 1, tp[i] = i;
jsort();
// Debug();
for (int w = 1, p = 0; p < len; M = p, w <<= 1)
{
//w:当前倍增的长度，w = x表示已经求出了长度为x的后缀的排名，现在要更新长度为2x的后缀的排名
//p表示不同的后缀的个数，很显然原字符串的后缀都是不同的，因此p = N时可以退出循环
p = 0;//这里的p仅仅是一个计数器000
for (int i = 1; i <= w; i++) tp[++p] = len - w + i;
for (int i = 1; i <= len; i++) if (sa[i] > w) tp[++p] = sa[i] - w; //这两句是后缀数组的核心部分，我已经画图说明
jsort();//此时我们已经更新出了第二关键字，利用上一轮的rak更新本轮的sa
swap(tp, rk);//这里原本tp已经没有用了
rk[sa[1]] = p = 1;
for (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;
//这里当两个后缀上一轮排名相同时本轮也相同
//Debug();
}
// for (int i = 1; i <= len; i++)
// printf("%d ", sa[i]);
}
//i号后缀：从i开始的后缀
//lcp(x,y):字符串x与字符串y的最长公共前缀，在这里指x号后缀与与y号后缀的最长公共前缀
int height[N];//lcp(sa[i],sa[i-1])，即排名为i的后缀与排名为i-1的后缀的最长公共前缀
int h[N];//height[rak[i]]，即i号后缀与它前一名的后缀的最长公共前缀
//性质：H[i]>=H[i-1]-1
void getHeight()
{
int j, k = 0;
for(int i = 1; i <= len; i++)
{
if(k) k--;
int j = sa[rk[i] - 1];
while(s[i + k] == s[j + k]) k++;
h[i]=height[rk[i]] = k;
//printf("%d\n", k);
}
}
/*
两个后缀的最大公共前缀lcp(x,y)=min(height[rank[x+1]~rank[y]])， 用rmq维护，O(1)查询
可重叠最长重复子串:height数组里的最大值
本质不同的子串的数量:枚举每一个后缀，第i个后缀对答案的贡献为len-sa[i]+1-height[i]
*/
/**********************主席树*********************/
int n, q, sz, num = 0;
int T[N];
int sum[N<<5], L[N<<5], R[N<<5];
#define mid (l+r)/2
inline int build(int l, int r)
{
int rt = ++ num;
sum[rt] = 0;
if (l < r)
{
L[rt] = build(l, mid);
R[rt] = build(mid+1, r);
}
return rt;
}
inline int update(int pre, int l, int r, int x)
{
int rt = ++ num;
L[rt] = L[pre];
R[rt] = R[pre];
sum[rt] = sum[pre]+1;
if (l < r)
{
if (x <= mid) L[rt] = update(L[pre], l, mid, x);
else R[rt] = update(R[pre], mid+1, r, x);
}
return rt;
}
inline int query(int u, int v, int l, int r, int k)
{
if (l >= r) return l;
int x = sum[L[v]] - sum[L[u]];
if (x >= k) return query(L[u], L[v], l, mid, k);
else return query(R[u], R[v], mid+1, r, k-x);
}
/*********************ST表**************/
int mm[N],dpMin[N][20];
//初始化Rmq,b数组下标从1开始，从0开始简单修改
void initRmq(int n,int b[]) //O(nlogn)预处理
{
mm[0]=-1;
for(int i=1;i<=n;i++)
{
mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
dpMin[i][0]=b[i];
}
for(int j=1;j<=mm[n];j++)
{
for(int i=1;i+(1<<j)-1<=n;i++)
{
dpMin[i][j]=min(dpMin[i][j-1],dpMin[i+(1<<(j-1))][j-1]);
}
}
}
int rmqMin(int x,int y)
{
int k=mm[y-x+1];
return min(dpMin[x][k],dpMin[y-(1<<k)+1][k]);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(sum,0,sizeof(sum));
memset(height,0,sizeof(height));
memset(dpMin,0,sizeof(dpMin));
memset(mm,0,sizeof(mm));
num=0;
int n,m;
scanf("%d%d",&n,&m);
scanf("%s",s+1);
len = strlen(s + 1);
suffixSort();
T[0]=build(1,n);
for(int i=1;i<=len;i++)
{
T[i]=update(T[i-1],1,n,sa[i]);
}
getHeight();
initRmq(n,height);
while(m--)
{
int x,y,k;
scanf("%d%d%d",&x,&y,&k);
int L=rk[x],R=rk[x];
int l=1,r=rk[x],g=y-x+1;
while(l<=r)
{
int h=(l+r)>>1;
if(h+1<=rk[x]&&rmqMin(h+1,rk[x])>=g)
{
r=h-1;
L=h;
// cout<<L<<" GG"<<endl;
}
else
l=h+1;
}
// cout<<L<<endl;
l=rk[x]+1,r=n;
while(l<=r)
{
int h=(l+r)>>1;
if(h>=rk[x]+1&&rmqMin(rk[x]+1,h)>=g)
{
l=h+1;
R=h;
}
else
r=h-1;
}
// printf("%d %d %d\n",rk[x],L,R);
if(R-L+1<k)
{
printf("-1\n");
}
else
{
printf("%d\n",query(T[L-1],T[R],1,n,k));
}
}
}
return 0;
}
/*
10
14 5
abcabcdabcddee
1 2 3
2 2 4
1 3 1
7 7 2
1 6 2
13 10
aabccaadeaaaa
*/