BZOJ 4278 [ONTAK2015]Tasowanie (后缀数组)

题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=4278

题解:

居然把后缀数组写成n^2的。。我真厉害。。

想了无数种方法，最后发现就是比后缀字典序排名，后缀数组即可

注意每个字符串的结尾处要加上一个\(\inf\), 因为相当于要把空位尽量后移使得非空位集中在前面

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 4e5+2;
const int S = 1e3+1;
int sa[N+3];
int rk[N+3];
int tmp[N+3];
int buc[N+3];
int a[N+3];
int ans[N+3];
int n,m;

void getSA()
{
	int *x = rk,*y = tmp;
	for(int i=0; i<=S; i++) buc[i] = 0;
	for(int i=1; i<=n; i++) buc[x[i]=a[i]]++;
	for(int i=1; i<=S; i++) buc[i] += buc[i-1];
	for(int i=n; i>=1; i--) sa[buc[x[i]]--] = i;
	int p = 0,s = S;
	for(int j=1; p<n; j<<=1)
	{
		p = 0;
		for(int i=n-j+1; i<=n; i++) {y[++p] = i;}
		for(int i=1; i<=n; i++) {if(sa[i]>j) y[++p] = sa[i]-j;}
		for(int i=1; i<=s; i++) buc[i] = 0;
		for(int i=1; i<=n; i++) buc[x[y[i]]]++;
		for(int i=1; i<=s; i++) buc[i] += buc[i-1];
		for(int i=n; i>=1; i--) sa[buc[x[y[i]]]--] = y[i];
		swap(x,y); p = 1; x[sa[1]] = 1;
		for(int i=2; i<=n; i++) x[sa[i]] = (y[sa[i]]==y[sa[i-1]] && y[sa[i]+j]==y[sa[i-1]+j]) ? p : ++p;
		s = p;
	}
	for(int i=1; i<=n; i++) rk[sa[i]] = i;
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%d",&a[i]); a[n+1] = S; n++;
	scanf("%d",&m);
	for(int i=n+1; i<=n+m; i++) scanf("%d",&a[i]); a[n+m+1] = S; m++;
	n += m;
	getSA();
	int i = 1,j = n-m+1,k = 1;
	while(i<n-m && j<n)
	{
		if(rk[i]<rk[j]) {ans[k] = a[i]; k++; i++;}
		else {ans[k] = a[j]; k++; j++;}
	}
	while(i<n-m) {ans[k] = a[i]; k++; i++;}
	while(j<n) {ans[k] = a[j]; k++; j++;}
	for(int i=1; i<n-1; i++) printf("%d ",ans[i]);
	return 0;
}