hdu 4898 The Revenge of the Princess’ Knight

时间:2024-01-11 14:21:26

传送阵:http://acm.hdu.edu.cn/showproblem.php?pid=4898

题目大意:一个首尾相连的字符串,将其分为k个子串,使得最大的字串最小

将所有子串排序,输出第k小即可

对于有循环节的串,用抽屉原理解决即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
int l;
char ch[];
}s[];
char ch[];
int vis[];
int n,k;
bool operator < (node a,node b){
for(int i=;i<=n;++i)
if(a.ch[i]!=b.ch[i])return a.ch[i]<b.ch[i];
return a.ch[]<b.ch[];
}
void work2(){
for(int i=;i<=n;++i)ch[i+n]=ch[i];
for(int i=;i<=n;++i){
int k=i+n;s[i].l=i;
for(int j=i,l=;j<k;++j,++l){
s[i].ch[l]=ch[j];
}
}
sort(s+,s+n+);
memset(vis,,sizeof(vis));
for(int i=;i<k;++i){
int l=s[i].l;
vis[l]=;vis[l+n]=;
}
for(int i=s[k].l;!vis[i];++i)
putchar(ch[i]);
return;
}
int Next[];
void get_pre(){
char s[];
for(int i=;i<=n;++i)s[i]=ch[i],s[i+n]=s[i];
int i=,j=,k=;
for(;i<=n&&j<=n&&k<=n;){
if(s[i+k]==s[j+k])k++;
else if(s[i+k]>s[j+k])i=i+k+,k=;
else if(s[i+k]<s[j+k])j=j+k+,k=;
if(i==j)j++;
}
if(i>j)i=j;
for(k=;k<=n;++k)
ch[k]=s[i+k-];
}
bool check(){
Next[]=Next[]=;
for(int i=;i<=n;++i){
int j=Next[i-];
while(j&&ch[j+]!=ch[i])j=Next[j];
Next[i]=ch[j+]==ch[i]?j+:;
}
int l=n-Next[n];
if(n%l==)return ;
return ;
}
void work1(){
int l=n-Next[n];
int a=n/l;
if(a<k)work2();
else{
if(a%k==){
int p=a/k;
int len=l*p;
for(int i=;i<=len;++i)putchar(ch[i]);
return;
}else{
int p=a/k+;
int len=l*p;
for(int i=;i<=len;++i)putchar(ch[i]);
return;
}
}
}
void init(){
scanf("%d%d",&n,&k);
scanf("%s",ch+);
get_pre();
}
void work(){
if(check())work1();
else work2();
}
int T;
int main(){
scanf("%d",&T);
while(T--){
init();
work();
puts("");
}
return ;
}