BZOJ 2882 & 后缀数组的傻逼实现

题意:

  一个字符环,求一个开头使字典序最小.

SOL:

  后缀数组打起来...然后居然卡过...10sec的实现我10936ms...居然卡过???

  rank倒三...啦啦啦啦啦....

  改个离散化会不会快点?....

Code:

  

/*==========================================================================
# Last modified: 2016-03-19 14:38
# Filename: 2882.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> #include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector> #define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1] #define maxn 700000
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459
#define INF 1070000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull; template<class T> inline
void read(T& num) {
bool start=false,neg=false;
char c;
num=0;
while((c=getchar())!=EOF) {
if(c=='-') start=neg=true;
else if(c>='0' && c<='9') {
start=true;
num=num*10+c-'0';
} else if(start) break;
}
if(neg) num=-num;
}
/*==================split line==================*/
int s[maxn];
int t[maxn],t2[maxn],c[maxn],sa[maxn];
int n; void build_sa(int m){
int *x=t,*y=t2;
FORP(i,0,m) c[i]=0;
FORP(i,0,n-1) c[x[i]=s[i]]++;
FORP(i,1,m) c[i]+=c[i-1];
FORM(i,n-1,0) sa[--c[x[i]]]=i;
for (int k=1;k<=n;k <<= 1){
int p=0;
FORP(i,n-k,n-1) y[p++]=i;
FORP(i,0,n-1) if (sa[i]>=k) y[p++]=sa[i]-k; FORP(i,0,m) c[i]=0;
FORP(i,0,n-1) c[x[y[i]]]++;
FORP(i,1,m) c[i]+=c[i-1];
FORM(i,n-1,0) sa[--c[x[y[i]]]]=y[i]; swap(x,y);
p=1; x[sa[0]]=0;
FORP(i,1,n-1)
if (y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k])
x[sa[i]]=p-1;
else x[sa[i]]=p++;
if (p>=n) break;
m=p;
}
}
int main(){
read(n); int tmp=n;
int m=0;
FORP(i,0,n-1) { read(s[i]); m=max(m,s[i]); s[n+i]=s[i];}
s[n+n]=m+1;
n=n+n+1;
build_sa(m+2);
for (int i=0;i<tmp;i++) printf("%d%c",s[(sa[0])%tmp+i],i==tmp-1?'\n':' ');
}
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