zoukankan      html  css  js  c++  java
  • kuangbin专题十六 KMP&&扩展KMP HDU3347 String Problem(最小最大表示法+kmp)

    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.OutputOutput 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


    只知道最小最大表示法。但是名次没思路。 后来看题解是循环节。。。。
    脑子真是废了。。 循环节用next数组搞定 即可

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<algorithm>
     4 using namespace std;
     5 const int maxn=2000010;
     6 char s[maxn],t[maxn];
     7 int Next[maxn],len;
     8 
     9 void prekmp(char* s) {
    10     int i,j;
    11     j=Next[0]=-1;
    12     i=0;
    13     while(i<len) {
    14         while(j!=-1&&s[i]!=s[j]) j=Next[j];
    15         if(s[++i]==s[++j]) Next[i]=Next[j];
    16         else Next[i]=j;
    17     }
    18 }
    19 
    20 int getmin(char* str) {
    21     int len2=strlen(str);
    22     int i=0,j=1,k=0;
    23     while(i<len&&j<len&&k<len) {
    24         int tmp=s[(i+k)%len2]-s[(j+k)%len2];
    25         if(!tmp) k++;
    26         else {
    27             if(tmp<0) j+=k+1;
    28             else i+=k+1;
    29             if(i==j) j++;
    30             k=0;
    31         }
    32     }
    33     return min(i,j);
    34 }
    35 
    36 int getmax(char* str) {
    37     int len2=strlen(str);
    38     int i=0,j=1,k=0;
    39     while(i<len&&j<len&&k<len) {
    40         int tmp=s[(i+k)%len2]-s[(j+k)%len2];
    41         if(!tmp) k++;
    42         else {
    43             if(tmp<0) i+=k+1;
    44             else j+=k+1;
    45             if(i==j) j++;
    46             k=0;
    47         }
    48     }
    49     return min(i,j);
    50 }
    51 
    52 int main() {
    53     while(~scanf("%s",s)) {
    54         len=strlen(s);
    55         prekmp(s);
    56         int num=1,len1=len-Next[len];
    57         if(len%len1==0)
    58             num=len/len1;
    59         strcpy(t,s);
    60         strcat(s,t);
    61         printf("%d %d %d %d
    ",getmin(s)+1,num,getmax(s)+1,num);
    62     }
    63 }


  • 相关阅读:
    30天内自动登录
    本地保存cookie
    thymeleaf 基本表达式
    适配器(adapter)与fragment之间、fragment与activity之间的通信问题
    String/Stringbuilder/StringBuffer
    if else与switch for与foreach
    iOS App 签名的原理
    iOS category内部实现原理
    iOS中的事件的产生和传递
    RunLoop
  • 原文地址:https://www.cnblogs.com/ACMerszl/p/10321817.html
Copyright © 2011-2022 走看看