P2870 [USACO07DEC]Best Cow Line G(后缀数组)

题目链接
题意：给你一个字符串，每次从首或尾取一个字符组成字符串，问所有能够组成的字符串中字典序最小的一个。
思路：首先定义两个指针L和R,如果s[L]<s[R],肯定选s[R],s[L]>s[R]，选s[R].对于s[L]==s[R]的情况,可以考虑L位置的rank以及R位置反串的rank值大小。所以还需要将s的反串构造进去，建立后缀数组。
突然发现其实就一种，因为如果s[L]<s[R]，那么rank[l]一定也小于反串的r的rank值。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define Max_N 1000100
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) 
int wa1[Max_N*3], wb1[Max_N*3], wv1[Max_N*3], ws1[Max_N*3]; 
int c0(int *r, int a, int b)  
{
	return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
}
 
int c12(int k, int *r, int a, int b) 
{
	if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1); 
    else return r[a] < r[b] || r[a] == r[b] && wv1[a+1] < wv1[b+1];
}
    
 void sort(int *r, int *a, int *b, int n, int m) 
 { 
	int i; 
	for (i = 0; i < n; i++) wv1[i] = r[a[i]]; 
	for (i = 0; i < m; i++) ws1[i] = 0; 
	for (i = 0; i < n; i++) ws1[wv1[i]]++; 
	for (i = 1; i < m; i++) ws1[i] += ws1[i-1]; 
	for (i = n-1; i >= 0; i--) b[--ws1[wv1[i]]] = a[i]; 
	return; 
 }
void dc3(int *r, int *sa, int n, int m)
{
	int i, j, *rn = r + n, *san = sa + n,ta = 0,tb = (n + 1) / 3,tbc = 0,p; 
    r[n] = r[n+1] = 0; 
    for (i = 0; i < n; i++) if (i % 3 != 0) wa1[tbc++] = i; 
    sort(r+2, wa1, wb1, tbc, m); 
    sort(r+1, wb1, wa1, tbc, m); 
    sort(r, wa1, wb1, tbc, m); 
    for (p = 1,rn[F(wb1[0])] = 0,i = 1;i < tbc; i++) 
    rn[F(wb1[i])] = c0(r, wb1[i-1], wb1[i]) ? p - 1 : p++; 
    if (p < tbc) dc3(rn, san, tbc, p); 
    else for (i = 0; i < tbc; i++) san[rn[i]] = i;
    for (i = 0; i < tbc; i++) if (san[i] < tb) wb1[ta++] = san[i] * 3; 
    if (n % 3 == 1) wb1[ta++] = n - 1; 
    sort(r, wb1, wa1, ta, m); 
    for (i = 0; i < tbc; i++) wv1[wb1[i] = G(san[i])] = i; 
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++) 
    sa[p] = c12(wb1[j] % 3, r, wa1[i], wb1[j]) ? wa1[i++] : wb1[j++]; 
    for(; i < ta; p++) sa[p] = wa1[i++]; 
    for(; j < tbc; p++) sa[p] = wb1[j++]; 
    return; 
}
// str sa 都要开三倍的；
void da(int str[], int sa[], int rank1[], int height1[], int n, int m)
{
	for (int i = n; i < n*3; i++)
		str[i] = 0;
	dc3(str, sa, n+1, m);
	int i, j, k = 0;
	for (i = 0; i <= n; i++) rank1[sa[i]] = i;
	for (i = 0; i < n; i++)
	{
		if (k) k--;
		j = sa[rank1[i] - 1];
		while (str[i+k] == str[j+k]) k++;
		height1[rank1[i]] = k;
	}
	return;
}
char s[Max_N];
int str[3*Max_N];
int sa[3*Max_N];//sa[i]表示将所有后缀排序后第i小的后缀的编号。
int rank1[3*Max_N];//rank1[i]表示后缀i的排名。
int height1[3*Max_N];
char ans[Max_N];
int main()
{
	/*char buf[100];
	cin >> buf;
	int n = strlen(buf);
	//cout << n << endl;
	int m = 128;
	for (int i = 0; i < strlen(buf); i++)
		str[i] = int(buf[i]);
	str[n] = 0;
	da(str, sa, rank1, height1, n, m);
	for (int i = 0; i <= n; i++)
		cout << sa[i] << endl;
	return 0;
	*/
	int n;
	scanf("%d",&n);
	int k=0;
	for(int i=0;i<n;i++)
	{
		char a[2];
		scanf("%s",a);
		s[k++]=a[0];
	}
	s[k]='#';
	for(int i=0;i<k;i++)
	{
		s[k+i+1]=s[k-i-1];
	}
	k=2*n+1;
	for(int i=0;i<k;i++)
	{
		str[i]=int(s[i]);
	}
	str[k]=0;
	int m=128;
	da(str, sa, rank1, height1, k, m);
	int L=0,R=n-1;
	int cnt=0;
	while(L<=R)
	{
		if(s[L]<s[R])
		{
			ans[cnt++]=s[L];
			L++;
		}
		else if(s[L]>s[R])
		{
			ans[cnt++]=s[R];
			R--;
		}
		else
		{
			int m=(n-R);
			if(rank1[L]>rank1[n+m])
			{
				ans[cnt++]=s[R];
				R--;
			}
			else
			{
				ans[cnt++]=s[L];
				L++;
			}
		}
	}
	ans[cnt]='\0';
	int tot=0;
	for(int i=0;i<cnt;i++)
	{
		printf("%c",ans[i]);
		tot++;
		if(tot%80==0)
		{
			printf("\n");
		}
	}
 
}

标签：Cow,wb1,++,int,P2870,wa1,tbc,sa,USACO07DEC
来源： https://www.cnblogs.com/2462478392Lee/p/13798961.html