hdu3374 String Problem【最小表示法】【exKMP】

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

 

题意：

找到给定的字符串$S$的最小表示法和最大表示法。并且找到他在所有循环同构串中的出现次数。

思路：

首先肯定是要先分别找出最大表示法和最小表示法对应的起始下标。

方法就是先复制一倍接在后面，然后用两个指针分别从0和1开始往后扫描，在第一个发现不相等的位置比较$s[i+k]$和$s[j+k]$的大小。

然后用exKMP分别对最小表示法和最大表示法求$extend$数组。

然后数一下有多少个$extend[i] = n$，这就是个数。

  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)//计算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)//第一种情况，可以直接得到next[i]的值
 34         nxt[i]=nxt[i-po];
 35         else//第二种情况，要继续匹配才能得到next[i]的值
 36         {
 37             j=nxt[po]+po-i;
 38             if(j<0)j=0;//如果i>po+next[po],则要从头开始匹配
 39             while(i+j<len&&str[j]==str[j+i])//计算next[i]
 40             j++;
 41             nxt[i]=j;
 42             po=i;//更新po的位置
 43         }
 44     }
 45 }
 46 //计算extend数组
 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);//计算子串的next数组
 51     while(s1[i]==s2[i]&&i<l2&&i<len)//计算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)//第一种情况，直接可以得到ex[i]的值
 58         ex[i]=nxt[i-po];
 59         else//第二种情况，要继续匹配才能得到ex[i]的值
 60         {
 61             j=ex[po]+po-i;
 62             if(j<0)j=0;//如果i>ex[po]+po则要从头开始匹配
 63             while(i+j<len&&j<l2&&s1[j+i]==s2[j])//计算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 }

 

posted @ 2018-11-20 15:37  wyboooo  阅读(249)  评论(0)  编辑  收藏  举报