HDU 3374 String Problem （字符串最小表示法+KMP）

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字符串处理：字符串最小表示法

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探讨了如何解决一个特定的字符串问题，即在给定一个长度不超过1e6的字符串时，找到其中字典序最小和最大的串，并计算它们在环状字符串中的出现次数。解决方案包括使用字符串的最小表示法确定最小、最大字典序的串，然后通过KMP算法进行匹配。

String Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4951 Accepted Submission(s): 2014

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

题意：

给你一个字符串(长度不超过1e6)，字符串是一个环，让你求其中字典序最小和最大的，并求其出现的次数（在整个环中）。

思路：

字符串的最小表示法求出最小、最大字典序的串，然后kmp跑就行了。

代码：

#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=2000010;
char str[maxn];
char s[maxn],p[maxn];
int len;
int n,m,nxt[maxn];
int cnt,as[maxn],ans;
void getfail()
{
    int i=0,j=-1;
    nxt[0]=-1;
    while(i<len)
    {
        if(j==-1||p[i]==p[j])
        {
            i++;j++;
            nxt[i]=j;
        }
        else j=nxt[j];
    }
}
void kmp(int &ans)
{
    int k,i,j;
    getfail();
    i=0;j=0;
    while(i<m)
    {
        if(j==-1||s[i]==p[j])
        {
            i++;j++;
        }
        else j=nxt[j];
        if(j==len) ans++;
    }
}
int find_pos()
{
    int p1=0,p2=1,k=0;
    while(p1<len&&p2<len&&k<len)
    {
        int tmp=str[(p1+k)%len]-str[(p2+k)%len];
        if(!tmp) k++;
        else
        {
            tmp>0?p1+=k+1:p2+=k+1;
            if(p1==p2) p2++;
            k=0;
        }
    }
    return min(p1,p2);
}
int find_pos2()
{
    int p1=0,p2=1,k=0;
    while(p1<len&&p2<len&&k<len)
    {
        int tmp=str[(p1+k)%len]-str[(p2+k)%len];
        if(!tmp) k++;
        else
        {
            tmp<0?p1+=k+1:p2+=k+1;
            if(p1==p2) p2++;
            k=0;
        }
    }
    return min(p1,p2);
}
void go()
{
    int id1,ans1=0,id2,ans2=0;
    int pos=find_pos();
    id1=pos+1;
    int l=0;
    for(int i=pos;i<len+pos;i++)
    {
        p[l++]=str[i%len];
    }
    p[len]='\0';
    kmp(ans1);
    pos=find_pos2();
    id2=pos+1;
    l=0;
    for(int i=pos;i<len+pos;i++)
    {
        p[l++]=str[i%len];
    }
    p[len]='\0';
    kmp(ans2);
    printf("%d %d %d %d\n",id1,ans1,id2,ans2);
}

int main()
{
    int t,i,cas=1;
    while(scanf("%s",s)!=EOF)
    {
        len=strlen(s);
        for(int i=0;i<len;i++)
        {
            str[i]=s[i];
            s[i+len]=s[i];
        }
        str[len]='\0';
        s[2*len-1]='\0';
        m=2*len-1;
        go();
    }
    return 0;
}