\(\color{#0066ff}{ 题目描述 }\)
给定一个字符串,求排名第k小的串
\(\color{#0066ff}{输入格式}\)
第一行给定主串(len<=90000)
第二行给定询问个数T<=500
随后给出T行T个询问,每次询问排名第k小的串,范围在int内
\(\color{#0066ff}{输出格式}\)
对于每一个询问,输出T行,每行为排名第k小的串
\(\color{#0066ff}{输入样例}\)
aaa
2
2
3
\(\color{#0066ff}{输出样例}\)
aa
aaa
\(\color{#0066ff}{数据范围与提示}\)
none
\(\color{#0066ff}{ 题解 }\)
求第k小,考虑在Sam上贪心
这时我们需要的siz是节点,所以统一dfs处理(注意记忆化)
然后依次贪心就行了
#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL in() {
char ch; int x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 2e5 + 5;
struct SAM {
protected:
struct node {
node *ch[26], *fa;
int len, siz;
node(int len = 0, int siz = 0): fa(NULL), len(len), siz(siz) {
memset(ch, 0, sizeof ch);
}
};
node *root, *tail, *lst;
node pool[maxn];
void extend(int c) {
node *o = new(tail++) node(lst->len + 1), *v = lst;
for(; v && !v->ch[c]; v = v->fa) v->ch[c] = o;
if(!v) o->fa = root;
else if(v->len + 1 == v->ch[c]->len) o->fa = v->ch[c];
else {
node *n = new(tail++) node(v->len + 1), *d = v->ch[c];
std::copy(d->ch, d->ch + 26, n->ch);
n->fa = d->fa, d->fa = o->fa = n;
for(; v && v->ch[c] == d; v = v->fa) v->ch[c] = n;
}
lst = o;
}
void clr() {
tail = pool;
root = lst = new(tail++) node();
}
void getans(int k, node *o) {
if(!k) return;
for(int i = 0; i <= 25; i++) {
if(o->ch[i]) {
if(o->ch[i]->siz >= k) {
putchar((char)(i + 'a'));
getans(k - 1, o->ch[i]);
return;
}
else k -= o->ch[i]->siz;
}
}
}
void getsiz(node *o) {
if(o->siz) return;
o->siz = 1;
for(int i = 0; i <= 25; i++) if(o->ch[i]) getsiz(o->ch[i]), o->siz += o->ch[i]->siz;
}
public:
SAM() { clr(); }
void ins(char *s) { for(char *p = s; *p; p++) extend(*p - 'a'); }
void getsiz() { getsiz(root); }
void getans(int k) { getans(k, root); }
}sam;
char s[maxn];
int main() {
scanf("%s", s);
sam.ins(s);
sam.getsiz();
for(int T = in(); T --> 0;) {
int k = in();
sam.getans(k);
puts("");
}
return 0;
}