PAT (Top Level) Practise 1005 Programming Pattern (35)

后缀数组。排序之后得到height数组，然后从上到下将height>=len的都分为一组，然后找到第一组个数最多的输出即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
inline int read()
{
    char c = getchar();  while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

const int maxn=1048576+10;

int wa[maxn],wb[maxn],wv[maxn],WS[maxn];
int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++) WS[i]=0;
    for(i=0; i<n; i++) WS[x[i]=r[i]]++;
    for(i=1; i<m; i++) WS[i]+=WS[i-1];
    for(i=n-1; i>=0; i--) sa[--WS[x[i]]]=i;
    for(j=1,p=1; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++) y[p++]=i;
        for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
        for(i=0; i<n; i++) wv[i]=x[y[i]];
        for(i=0; i<m; i++) WS[i]=0;
        for(i=0; i<n; i++) WS[wv[i]]++;
        for(i=1; i<m; i++) WS[i]+=WS[i-1];
        for(i=n-1; i>=0; i--) sa[--WS[wv[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
    return;
}

int Rank[maxn],height[maxn];
void calheight(int *r,int *sa,int n)
{
    int i,j,k=0;
    for(i=1; i<=n; i++) Rank[sa[i]]=i;
    for(i=0; i<n; height[Rank[i++]]=k)
        for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);
    return;
}

int T,n,a[maxn],SA[maxn],sum[maxn];
char str[maxn],op[5];

int Find(int D,int l,int r)
{
    int pos=-1;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(sum[mid]-D>1) r=mid-1;
        else if(sum[mid]-D==1) pos=mid,r=mid-1;
        else l=mid+1;
    }
    return pos;
}

int main()
{
    int len; scanf("%d",&len); getchar();
    gets(str); n=strlen(str);
    for(int i=0;i<n;i++) a[i]=(int)str[i];
    a[n]=0; da(a,SA,n+1,300); calheight(a,SA,n);
    int L=1,R=1;
    int ansL,ansR,ans=0;
    while(1)
    {
        if(height[R]>=len)
        {
            R++;
            if(R-L+1>ans) ans=R-L+1, ansL = L, ansR = R;
        }
        else L=R+1, R=L;
        if(L>n||R>n) break;
    }
    for(int i=SA[ansL];i<=SA[ansL]+len-1;i++) printf("%c",str[i]);
    printf(" %d\n",ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/zufezzt/p/5744413.html