• 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?'
    ':' ');
    }
    
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  • 原文地址:https://www.cnblogs.com/YCuangWhen/p/5295312.html
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