一道后缀数组的模版题

Part 1:

后缀数组

做法:

我觉得一张图能很明确的表示..

这样就能开心的求出rank数组了

看图说话..

http://user.qzone.qq.com/664045109/blog/1388323241

http://user.qzone.qq.com/664045109/blog/1430923906

Part 2:

LCP

http://blog.csdn.net/shiqi_614/article/details/7982915

首先,我们要知道一个结论

我们设h[i]代表第i个子串与第b[rank[i]-1].id的最长公共前缀

那么h[i]>=h[i-1]-1

这么思考:

i和b[rank[i]-1].id都去掉它的第一位..(以下把后者简写为b[].id)

得到的东西的前面h[i-1]-1位和b[].id是一样的

而b[].id去掉第一位后得到的即是b[].id+1

那么lcp(i+1,b[].id+1)=h[i-1]-1

根据lcp的奇怪的性质

lcp(rank[i],rank[j]) = min(lcp(rank[i],rank[k]),lcp(rank[k],rank[j]))

我们可以发现它的答案至少是h[i-1]-1

代码:

Past:

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<math.h>
#include<time.h>
#include<vector>
#include<bitset>
#include<memory>
#include<utility>
#include<fstream>
#include<stdio.h>
#include<sstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[100005];
struct node
{
    int x;
    int y;
    int id;
    friend bool operator < (const node &a,const node &b)
    {
        if (a.x==b.x)
        {
            return a.y<b.y;
        }
        return a.x<b.x;
    }
};
node b[100005];
int rank[100005];
int n;
void suffix_sort(int x)
{
    int i;
    for (i=0;i<n;i++)
    {
        b[i].x=rank[b[i].id];
        if (b[i].id+x<n)
        {
            b[i].y=rank[b[i].id+x];
        }
        else
        {
            b[i].y=-1;
        }
    }
    sort(b,b+n);
    rank[b[0].id]=0;
    for (i=1;i<n;i++)
    {
        rank[b[i].id]=rank[b[i-1].id];
        if (b[i-1]<b[i])
        {
            rank[b[i].id]++;
        }
    }
    if (x>n) return;
    suffix_sort(x*2);
}
int c[100005];
int ans[100005];
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    gets(a);
    n=strlen(a);
    int i;
    for (i=0;i<n;i++)
    {
        b[i].x=a[i];
        b[i].id=i;
    }
    sort(b,b+n);
    rank[b[0].id]=0;
    for (i=1;i<n;i++)
    {
        rank[b[i].id]=rank[b[i-1].id];
        if (b[i-1]<b[i])
        {
            rank[b[i].id]++;
        }
    }
    suffix_sort(1);
    for (i=0;i<n;i++)
    {
        printf("%d ",b[i].id+1);
    }
    putchar(10);
    int now=0;
    for (i=0;i<n;i++)
    {
        if (rank[i]==n-1)
        {
            now=0;
            continue;
        }
        int t=b[rank[i]+1].id;
        now--;
        if (now<0) now=0;
        for (;;)
        {
            if (a[now+t]==a[now+i])
            {
                now++;
            }
            else
            {
                break;
            }
        }
        ans[rank[i]]=now;
    }
    for (i=0;i<n-1;i++)
    {
        printf("%d ",ans[i]);
    }
    return 0;
}

Now:

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<time.h>
#include<bitset>
#include<vector>
#include<memory>
#include<utility>
#include<stdio.h>
#include<sstream>
#include<fstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[100005];
struct node
{
    int x;
    int y;
    int id;
    friend bool operator < (const node &a,const node &b)
    {
        if (a.x==b.x) return a.y<b.y;
        return a.x<b.x;
    }
};
node b[100005];
int n;
int c[100005];
int lcp[100005];
void preffix_sort(int x=1)
{
    if (x>n) return;
    int i;
    for (i=0;i<n;i++)
    {
        if (b[i].id+x<n)
        {
            b[i].y=c[b[i].id+x];
        }
        else
        {
            b[i].y=-1;
        }
    }
    sort(b,b+n);
    for (i=0;i<n;i++)
    {
        if ((i==0)||(b[i-1]<b[i]))
        {
            c[b[i].id]=i;
        }
        else
        {
            c[b[i].id]=c[b[i-1].id];
        }
    }
    for (i=0;i<n;i++)
    {
        b[i].x=c[b[i].id];
    }
    preffix_sort(x*2);
}
int ans[100005];
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    scanf("%s",a);
    int i;
    for (i=0;a[i]!='\0';i++)
    {
        b[i].x=a[i];
        b[i].y=0;
        b[i].id=i;
    }
    n=i;
    sort(b,b+n);
    for (i=0;i<n;i++)
    {
        c[b[i].id]=b[i].x;
    }
    preffix_sort();
    for (i=0;i<n;i++)
    {
        printf("%d ",b[i].id+1);
    }
    printf("\n");
    int now=0;
    for (i=0;i<n;i++)
    {
        if (c[i]==0)
        {
            now=0;
            continue;
        }
        now--;
        if (now<0) now=0;
        int t=b[c[i]-1].id;
        for (;;)
        {
            if (a[i+now]==a[t+now])
            {
                now++;
            }
            else
            {
                break;
            }
        }
        ans[c[i]]=now;
    }
    for (i=1;i<n;i++)
    {
        printf("%d ",ans[i]);
    }
    return 0;
}