output field can mis sep pan 題意 easy example

String Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4828 Accepted Submission(s): 1949

Problem Description 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.

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

Output 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.

Sample Input abcder aaaaaa ababab

Sample Output 1 1 6 1 1 6 1 6 1 3 2 3

Author WhereIsHeroFrom

Source HDOJ Monthly Contest – 2010.04.04

題意：

找到給定的字符串$S$的最小表示法和最大表示法。並且找到他在所有循環同構串中的出現次數。

思路：

首先肯定是要先分別找出最大表示法和最小表示法對應的起始下標。

方法就是先復制一倍接在後面，然後用兩個指針分別從0和1開始往後掃描，在第一個發現不相等的位置比較$s[i+k]$和$s[j+k]$的大小。

然後用exKMP分別對最小表示法和最大表示法求$extend$數組。

然後數一下有多少個$extend[i] = n$，這就是個數。

  1 #include<iostream>
  2 #include<bits/stdc++.h>
  3 #include<cstdio>
  4 #include<cmath>
  5 //#include<cstdlib>
  6 #include<cstring>
  7 #include<algorithm>
  8 //#include<queue>
  9 #include<vector>
 10 //#include<set>
 11 //#include<climits>
 12 //#include<map>
 13 using namespace std;
 14 typedef long long LL;
 15 #define N 100010
 16 #define pi 3.1415926535
 17 #define inf 0x3f3f3f3f
 18 
 19 const int maxn = 1e6 + 6;
 20 char s[maxn * 2], t[maxn];
 21 int nxt[maxn * 2], ex_min[maxn * 2], ex_max[maxn * 2];
 22 
 23 void GETNEXT(char *str)
 24 {
 25     int i=0,j,po,len=strlen(str);
 26     nxt[0]=len;//初始化next[0]
 27     while(str[i]==str[i+1]&&i+1<len)//計算next[1]
 28     i++;
 29     nxt[1]=i;
 30     po=1;//初始化po的位置
 31     for(i=2;i<len;i++)
 32     {
 33         if(nxt[i-po]+i<nxt[po]+po)//第一種情況，可以直接得到next[i]的值
 34         nxt[i]=nxt[i-po];
 35         else//第二種情況，要繼續匹配才能得到next[i]的值
 36         {
 37             j=nxt[po]+po-i;
 38             if(j<0)j=0;//如果i>po+next[po],則要從頭開始匹配
 39             while(i+j<len&&str[j]==str[j+i])//計算next[i]
 40             j++;
 41             nxt[i]=j;
 42             po=i;//更新po的位置
 43         }
 44     }
 45 }
 46 //計算extend數組
 47 void EXKMP(char *s1,char *s2, int *ex)
 48 {
 49     int i=0,j,po,len=strlen(s1),l2=strlen(s2);
 50     GETNEXT(s2);//計算子串的next數組
 51     while(s1[i]==s2[i]&&i<l2&&i<len)//計算ex[0]
 52     i++;
 53     ex[0]=i;
 54     po=0;//初始化po的位置
 55     for(i=1;i<len;i++)
 56     {
 57         if(nxt[i-po]+i<ex[po]+po)//第一種情況，直接可以得到ex[i]的值
 58         ex[i]=nxt[i-po];
 59         else//第二種情況，要繼續匹配才能得到ex[i]的值
 60         {
 61             j=ex[po]+po-i;
 62             if(j<0)j=0;//如果i>ex[po]+po則要從頭開始匹配
 63             while(i+j<len&&j<l2&&s1[j+i]==s2[j])//計算ex[i]
 64             j++;
 65             ex[i]=j;
 66             po=i;//更新po的位置
 67         }
 68     }
 69 }
 70 
 71 int main()
 72 {
 73     while(scanf("%s", s) != EOF){
 74         int n = strlen(s);
 75         for(int i = 0; i < n; i++)s[n + i] = s[i];
 76         int i = 0, j = 1, k;
 77         while(i < n && j < n){
 78             for(k = 0; k < n && s[i + k] == s[j + k]; k++);
 79             if(k == n)break;
 80             if(s[i + k] > s[j + k]){
 81                 i = i + k + 1;
 82                 if(i == j)i++;
 83             }
 84             else{
 85                 j = j + k + 1;
 86                 if(i == j)j++;
 87             }
 88         }
 89         int rnk_min = min(i, j);
 90 
 91         i = 0, j = 1;
 92         while(i < n && j < n){
 93             for(k = 0; k < n && s[i + k] == s[j + k]; k++);
 94             if(k == n)break;
 95             if(s[i + k] > s[j + k]){
 96                 j = j + k + 1;
 97                 if(i == j)j++;
 98             }
 99             else{
100                 i = i + k + 1;
101                 if(i == j)i++;
102             }
103         }
104         int rnk_max = min(i, j);
105 
106         int min_cnt = 0;
107         memcpy(t, s + rnk_min, n);
108         EXKMP(s, t, ex_min);
109         for(int i = 0;i < n; i++){
110             if(ex_min[i] == n)min_cnt++;
111         }
112         int max_cnt = 0;
113         memcpy(t, s + rnk_max, n);
114         EXKMP(s, t, ex_max);
115         for(int i = 0; i < n; i++){
116             if(ex_max[i] == n)max_cnt++;
117         }
118         printf("%d %d %d %d\n", rnk_min + 1, min_cnt, rnk_max + 1, max_cnt);
119     }
120     return 0;
121 }

hdu3374 String Problem【最小表示法】【exKMP】