解題思路:

建個字尾自動機,然後在樹上跑前K個字元的dfs,直到遇到樹上沒有的即停止.

演算法複雜度O(n*m)

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int mx = 2e4+10;
int n,m,K,cnt;
char str[mx],ans[mx];
struct sam{
	int ch[mx][26];
	int f[mx];
	int len[mx];
	int c[mx];
	int id[mx];
	int sz,rt,last;
	int newnode(int v){
		len[++sz] = v;
		memset(ch[sz],0,sizeof(ch[sz]));
		return sz;
	}
	void init(){
		sz = 0;
		rt = last = newnode(0);
	}
	void insert(int c){
		int p = last,np = newnode(len[last]+1);
		while(p&&!ch[p][c]) ch[p][c] = np,p = f[p];
		if(!p) f[np] = rt;
		else{
			int q = ch[p][c];
			if(len[q]==len[p]+1) f[np] = q;
			else{
				int nq = newnode(len[p]+1);
				memcpy(ch[nq],ch[q],sizeof(ch[q]));
				f[nq] = f[q];
				f[q] = f[np] = nq;
				while(p&&ch[p][c]==q) ch[p][c] = nq,p = f[p];
			}
		}
		last = np;
	}
	bool dfs(int p){
		if(cnt==m) return 0;
		for(int i=K;~i;i--){
			if(!ch[p][i]){
				ans[cnt++] = i+'a';
				return 1;
			}
			ans[cnt++] = i+'a';
			if(dfs(ch[p][i])) return 1;
			cnt--;
		}
		return 0;
	}
}word;
int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d%d",&n,&m,&K);
		K--,cnt = 0;
		scanf("%s",str); 
		word.init();
		for(int i=0;i<n;i++) word.insert(str[i]-'a');
		word.dfs(word.rt);ans[cnt] = 0;
		puts(ans);
	}
	return 0;
}