HDOJ3374 String Problem 【KMP最小循环节点】+【最小(大)表示法】

 

String Problem

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1442    Accepted Submission(s): 645
[/b]

[align=left]Problem Description[/align]
Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:

String Rank

SKYLONG 1

KYLONGS 2

YLONGSK 3

LONGSKY 4

ONGSKYL 5

NGSKYLO 6

GSKYLON 7

and lexicographically first of them is GSKYLON, lexicographically last is YLONGSK, both of them appear only once.

  Your task is easy, calculate the lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), its times, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.

 

 

[align=left]Input[/align]
  Each line contains one line the string S with length N (N <= 1000000) formed by lower case letters.
 

 

[align=left]Output[/align]
Output four integers separated by one space, lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), the string’s times in the N generated strings, lexicographically last string’s Rank (if there
are multiple answers, choose the smallest one), and its times also.
 

 

[align=left]Sample Input[/align]

abcder
aaaaaa
ababab

 

 

[align=left]Sample Output[/align]

1 1 6 1
1 6 1 6
1 3 2 3

 

 

 

#include <stdio.h>
#include <string.h>
#define maxn 1000002
char str[maxn];
int len, next[maxn];

void getNext(){
int i = 0, j = -1;
next[0] = -1;
while(i < len){
if(j == -1 || str[i] == str[j]){
++i; ++j;
next[i] = j;
}else j = next[j];
}
}

int minR(){
int i = 0, j = 1, k = 0, t;
while(i < len && j < len && k < len){
t = str[i+k >= len ? i+k-len : i+k] -
str[j+k >= len ? j+k-len : j+k];
if(t == 0) ++k;
else{
if(t > 0) i += k + 1;
else j += k + 1;
k = 0;
if(i == j) ++j;
}
}
return i < j ? i : j;
}

int maxR(){
int i = 0, j = 1, k = 0, t;
while(i < len && j < len && k < len){
t = str[i+k >= len ? i+k-len : i+k] -
str[j+k >= len ? j+k-len : j+k];
if(t == 0) ++k;
else{
if(t > 0) j += k + 1;
else i += k + 1;
k = 0;
if(i == j) ++j;
}
}
return i < j ? i : j;
}

int main(){
int minlen, num;
while(scanf("%s", str) == 1){
len = strlen(str);
getNext();
if(len % (len - next[len]) == 0) //最小循环节点
minlen = len - next[len];
else minlen = len;
num = len / minlen;
printf("%d %d %d %d\n", minR()+1, num, maxR()+1, num);
}
return 0;
}