【BZOJ1031】【JSOI2007】字符串加密

一道很简单的后缀数组题

把字符串重复两次求个后缀数组，然后依次把开头在第一遍字符串中的字符串对应的最后末尾输出。

// <1031.cpp> - 05/14/16 22:00:59
// This file is created by XuYike's black technology automatically.
// Copyright (C) 2015 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long lol;
int gi(){
    int res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();
    if(ch=='-')fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
const int MAXN=200010;
const int INF=1e9;
int n,L;
char s[MAXN];
int sa[MAXN],rk[MAXN];
int v[MAXN],x[MAXN],y[MAXN];
void go(){
    int m=300;
    for(int i=1;i<=n;i++)v[x[i]=s[i]]++;
    for(int i=2;i<=m;i++)v[i]+=v[i-1];
    for(int i=n;i;i--)sa[v[x[i]]--]=i;
    for(int j=1;j<=n;j<<=1){
        int 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<=m;i++)v[i]=0;
        for(int i=1;i<=n;i++)v[x[y[i]]]++;
        for(int i=2;i<=m;i++)v[i]+=v[i-1];
        for(int i=n;i;i--)sa[v[x[y[i]]]--]=y[i];
        swap(x,y);x[sa[1]]=1;
        p=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;
        if(p==n)break;m=p;
    }
}
int main(){
    scanf("%s",s+1);
    L=strlen(s+1);
    for(int i=1;i<=L;i++)s[L+i]=s[i];
    n=L<<1;
    go();
    for(int i=1;i<=n;i++)if(sa[i]<=L)putchar(s[sa[i]+L-1]);
    return 0;
}

// <1031.cpp> - 05/14/16 22:00:59

// This file is created by XuYike's black technology automatically.

// I don't know what this program is.

#include <iostream>

#include <vector>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <cmath>

using namespace std;

typedef long long lol;

int gi(){

int res=0,fh=1;char ch=getchar();

while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();

if(ch=='-')fh=-1,ch=getchar();

while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();

return fh*res;

}

const int MAXN=200010;

const int INF=1e9;

int n,L;

char s[MAXN];

int sa[MAXN],rk[MAXN];

int v[MAXN],x[MAXN],y[MAXN];

void go(){

int m=300;

for(int i=1;i<=n;i++)v[x[i]=s[i]]++;

for(int i=2;i<=m;i++)v[i]+=v[i-1];

for(int i=n;i;i--)sa[v[x[i]]--]=i;

for(int j=1;j<=n;j<<=1){

int 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<=m;i++)v[i]=0;

for(int i=1;i<=n;i++)v[x[y[i]]]++;

for(int i=2;i<=m;i++)v[i]+=v[i-1];

for(int i=n;i;i--)sa[v[x[y[i]]]--]=y[i];

swap(x,y);x[sa[1]]=1;

p=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;

if(p==n)break;m=p;

}

int main(){

scanf("%s",s+1);

L=strlen(s+1);

for(int i=1;i<=L;i++)s[L+i]=s[i];

n=L<<1;

go();

for(int i=1;i<=n;i++)if(sa[i]<=L)putchar(s[sa[i]+L-1]);

return 0;

}