hdu5442Favorite Donut(好题)

Favorite Donut

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1961    Accepted Submission(s): 482

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 ofn 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 thei−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 are2n 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 ofn 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 (from1 to n) and the direction (0 for clockwise and 1 for counterclockwise).

 

Sample Input

24abab4aaab

 

Sample Output

2 04 0

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

方法：倍长字符串长度，构建后缀数组。

注意：顺时针时在字符串末尾补0(无穷小）;逆时针时在字符串末尾补'z'+1（无穷大）.

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;const int maxn=40000+100;char s[maxn],s1[maxn],s2[maxn];char s3[maxn],s4[maxn];int rank1[maxn];int sa[maxn],t[maxn],t2[maxn],c[maxn],n;//构造字符串s的后缀数组,每个字符值必须为0~m-1void build_sa(int m){    int *x=t,*y=t2;    //基数排序    for(int i=0;i<m;i++)    c[i]=0;    for(int i=0;i<n;i++)    c[x[i]=s[i]]++;    for(int i=1;i<m;i++)    c[i]+=c[i-1];    for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;    for(int k=1;k<=n;k<<=1){        int p=0;        //直接利用sa数组排序第二关键字        for(int i=n-k;i<n;i++)  y[p++]=i;        for(int i=0;i<n;i++)    if(sa[i]>=k)    y[p++]=sa[i]-k;        //基数排序第一关键字        for(int i=0;i<m;i++)    c[i]=0;        for(int i=0;i<n;i++)    c[x[y[i]]]++;        for(int i=1;i<m;i++)    c[i]+=c[i-1];        for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];        //根据sa和y计算新的x数组        swap(x,y);        p=1;        x[sa[0]]=0;        for(int i=1;i<n;i++)            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;        if(p>=n)            break;        m=p;                //下次基数排序的最大值    }    for(int i=0;i<n;i++)        rank1[i]=x[i];}int main(){    int t;    //freopen("5442.in","r",stdin);    scanf("%d",&t);    while(t--){        scanf("%d",&n);        int m=n;        scanf("%s",s);        for(int i=0;i<n;i++)            s[i+n]=s[i];        s[2*n]='0';        n=n*2+1;        build_sa(240);        int x,y;        x=0;        for(int i=1;i<m;i++)            if(rank1[i]>rank1[x])                x=i;        strcpy(s2,s);        int cnt=0;        for(int i=n-2;i>=0;i--)            s[cnt++]=s2[i];        s[cnt]=char('z'+1);        build_sa(240);        y=m-1;        for(int i=m-2;i>=0;i--)            if(rank1[i]>rank1[y])                y=i;        int z=y;        y=m-1-y;        cnt=0;        for(int i=x;;i++){            s3[cnt++]=s2[i];            if(cnt>=m)                break;        }        s3[cnt]='\0';        cnt=0;        for(int i=z;;i++){            s4[cnt++]=s[i];            if(cnt>=m)                break;        }        s4[cnt]='\0';        if(strcmp(s3,s4)==0){       if(x<=y)            printf("%d 0\n",x+1);        else        printf("%d 1\n",y+1);        }        else if(strcmp(s3,s4)<0)            printf("%d 1\n",y+1);        else        printf("%d 0\n",x+1);    }    return 0;}