HDU 5442——Favorite Donut——————【最大表示法+kmp | 后缀数组】

Favorite Donut

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1702    Accepted Submission(s): 430

Problem Description

Lulu has a sweet tooth. Her favorite food is ring donut. Everyday she buys a ring donut from the same bakery. A ring donut is consists of n parts. Every part has its own sugariness that can be expressed by a letter from a to z (from low to high), and a ring donut can be expressed by a string whose i-th character represents the sugariness of the i−th part in clockwise order. Note that z is the sweetest, and two parts are equally sweet if they have the same sugariness.

Once Lulu eats a part of the donut, she must continue to eat its uneaten adjacent part until all parts are eaten. Therefore, she has to eat either clockwise or counter-clockwise after her first bite, and there are 2n ways to eat the ring donut of n parts. For example, Lulu has 6 ways to eat a ring donut abc: abc,bca,cab,acb,bac,cba. Lulu likes eating the sweetest part first, so she actually prefer the way of the greatest lexicographic order. If there are two or more lexicographic maxima, then she will prefer the way whose starting part has the minimum index in clockwise order. If two ways start at the same part, then she will prefer eating the donut in clockwise order. Please compute the way to eat the donut she likes most.

 

Input

First line contain one integer T,T≤20, which means the number of test case.

For each test case, the first line contains one integer n,n≤20000, which represents how many parts the ring donut has. The next line contains a string consisted of n lowercase alphabets representing the ring donut.

 

Output

You should print one line for each test case, consisted of two integers, which represents the starting point (from 1 to n) and the direction (0 for clockwise and 1 for counterclockwise).

 

Sample Input

2

4

abab

4
aaab

 

Sample Output

2 0

4 0

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

 

题目大意：顺时针或逆时针遍历字符串，让求最大字典序。如果最大字典序唯一，那么输出得到最大字典序的位置及遍历顺序。如果不唯一，那么位置小的优先，如果位置也相同，那么顺时针优先。结果输出位置及遍历顺序。顺时针为0，逆时针为1。

 

吐槽：自己被自己玩死了。数据范围自己开小了，题目是2W，自己yy的1W，错到死　　WA。

 

解题思路：对于顺时针得到最大字典序的最小下标，可以用最大表示法直接得到。对于逆时针的最小下标，我们首先将原字符串反转，然后用最大表示法得到下标，但是得到的这个下标，并不是原串逆序得到最大字典序的最小下标，而是最大的。所以我们将反转后的串复制在另外一个空串中，再在尾部连上一个反转后的串，将这个长度为2n的串作为文本串。然后我们在反转后的串中提取出在原串中逆序最大字典序的串作为模式串，然后用kmp匹配，这时候匹配需要得到最大的匹配位置（由于是反转后的串，所以最大即在原串中最小）。最后先比较正序和逆序得到的字符串是否相同，不同取大的遍历顺序；相同则比较位置大小，小的或等的顺时针优先。

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+500;
char str[maxn],sp[maxn],revs[maxn],rvsp[maxn];
char tx[3*maxn];
int f[maxn];
int get_max(char *s){   //最大（最小）表示法得到的是正序遍历最大（最小）字典序开始位置。
    int i=0,j=1,k=0;
    int len=strlen(s);
    while(i<len&&j<len){
        if(k==len){
            break;
        }
        if(s[(i+k)%len]<s[(j+k)%len]){
            i=i+k+1>j? i+k+1:j+1;
            k=0;
        }else if(s[(i+k)%len]>s[(j+k)%len]){
            j=j+k+1>i? j+k+1:i+1;
            k=0;
        }else{
            k++;
        }
    }
    return min(i,j);
}
void get_s_p(char *sr,char *s,int st,int len){  //提取字符串
    for(int i=st,k=0;k<len;i++,k++){
        sr[i-st]=s[i%len];
    }
    sr[len]='\0';
}
void rev(char *s,int len){  //反转
    for(int i=len-1;i>=0;i--){
        revs[len-1-i]=s[i];
    } revs[len]='\0';
}

void getfail(char *P,int *F){
    int m=strlen(P);
    F[0]=F[1]=0;    int j;
    for(int i=1;i<m;i++){
        j=F[i];
        while(j&&P[i]!=P[j]) j=F[j];
        F[i+1]=P[i]==P[j]?j+1:0;
    }
}
int kmp(char *T,char *P){
    int n=strlen(T),m=strlen(P);
    int j=0;
    int ret=0;
    for(int i=0;i<n-1;i++){
        while(j&&P[j]!=T[i]) j=f[j];
        if(P[j]==T[i]) j++;
        if(j==m){
            ret=max(ret,i-m+1);     //对于反转后的串，需得到最大位置
            j=f[j];
        }
    }
    return ret;
}
int main(){
    int t , n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%s",str);
        rev(str,n);

        int ckw=get_max(str);
        get_s_p(sp,str,ckw,n);

        int ctckw=get_max(revs);
        get_s_p(rvsp,revs,ctckw,n);

        strcpy(tx,revs);
        strcat(tx,revs);

        getfail(rvsp,f);
        int ans=kmp(tx,rvsp);   //在反转后的串中的位置
        ans=n-1-ans;    //转化为原串中的位置
        int d=strcmp(sp,rvsp);
        if(d>0){
            printf("%d 0\n",ckw+1);
        }else if(d<0){
            printf("%d 1\n",ans+1);
        }else{
            if(ckw<=ans){
                printf("%d 0\n",ckw+1);
            }else{
                printf("%d 1\n",ans+1);
            }
        }
    }
    return 0;
}