HDU 3374 (KMP 最小表示法)

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2442    Accepted Submission(s): 1029

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:求循环串的字典序最大和最小的子串出现的次数和下标.

次数只要拿原串匹配一下统计就好了,下标用最小表示法找.

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define maxn 1111111

char P[maxn], T[maxn<<1];
int n, m;
#define next Next
int next[maxn];

void get_next (char *p) {
	int t;
	t = next[0] = -1;
	int j = 0, m = strlen (p);
	while (j < m) {
		if (t < 0 || p[j] == p[t]) {//匹配
			j++, t++;
			next[j] = t;
		}
		else //失配
			t = next[t];
	}
}

int kmp () {
	get_next (P);
	int i = 0, j = 0, ans = 0;
	while (i < n && j < m) {
		if (j < 0 || T[i] == P[j]) {
			i++, j++;
		}
		else {
			j = next[j];
		}
		if (j == m) {
            ans++;
            j = next[j];
		}
	}
	return ans;
}

int min_max_express (char *s, int len,bool flag) {
    int i=0,j=1,k=0;
    while (i<len && j<len && k<len) {
        int t = s[(j+k)%len]-s[(i+k)%len];
        if (t==0) k++;
        else {
            if (flag) {
                if (t>0) j+=k+1;
                else i+=k+1;
            }
            else {
                if (t>0) i+=k+1;
                else j+=k+1;
            }
            if (i==j) j++;
            k=0;
        }
    }
    return min(i,j);
}

int main () {
    //freopen ("in.txt", "r", stdin);
	while (scanf ("%s", P) == 1) {
        m = n = strlen (P);
        for (int i = 0; i < 2*n-1; i++) T[i] = P[i%n];
        n <<= 1; n--;
        T[n] = 0;
        int ans = kmp ();
        printf ("%d %d %d %d\n", min_max_express (P, m, 1)+1, ans, min_max_express (P, m, 0)+1, ans);
	}
	return 0;
}