hdu 5.2.3 boring counting

這道題從去年看到今年。。後綴數組什麼的果然是一個我的智商難以達到的地方。。暫時擱置，太考驗承受能力了。。

Boring counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 67 Accepted Submission(s): 21

Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

 

Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).

 

Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.

 

Sample Input

aaaa
ababcabb
aaaaaa
#

 

Sample Output

2
3
3

 

直接粘代碼了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 1200
char s[N];
int n, sa[4*N], rank[N], height[N];
int buf[4*N], ct[N], sx[N], sax[N];
inline bool leq(int a, int b, int x, int y)
{
    return (a < x || a == x && b <= y);
}
inline bool leq(int a, int b, int c, int x, int y, int z)
{
    return (a < x || a == x && leq(b, c, y, z));
}
inline int geti(int t, int nx, int sa[])
{
    return (sa[t]<nx ? sa[t]*3+1 : (sa[t]-nx)*3+2);
}
static void radix(int a[], int b[], int s[], int n, int k)
{ // sort a[0..n-1] to b[0..n-1] with keys in 0..k from s
    int i, t, sum;
    memset(ct, 0, (k + 1) * sizeof(int));
    for (i = 0; i < n; ++i) ct[s[a[i]]]++;
    for (i = 0, sum = 0; i <= k; ++i)
    {
        t = ct[i]; ct[i] = sum; sum += t;
    }
    for (i = 0; i < n; i++) b[ct[s[a[i]]]++] = a[i];
}
void suffix(int s[], int sa[], int n, int k)
{ // !!! require s[n] = s[n+1] = s[n+2] = 0, n >= 2.
    int i, j, e, p, t;
    int name = 0, cx = -1, cy = -1, cz = -1;
    int nx = (n+2)/3, ny = (n+1)/3, nz = n/3, nxz = nx+nz;
    int *syz = s + n + 3, *sayz = sa + n + 3;
    for (i=0, j=0; i < n + (nx - ny); i++)
    if (i%3 != 0) syz[j++] = i;
    radix(syz , sayz, s+2, nxz, k);
    radix(sayz, syz , s+1, nxz, k);
    radix(syz , sayz, s , nxz, k);
    for (i = 0; i < nxz; i++)
    {
        if (s[ sayz[i] ] != cx || s[ sayz[i] + 1 ] != cy ||s[ sayz[i] + 2 ] != cz)
        {
            name++; cx = s[ sayz[i] ];
            cy = s[ sayz[i] + 1 ]; cz = s[ sayz[i] + 2 ];
        }
        if (sayz[i] % 3 == 1) syz[ sayz[i] / 3 ] = name;
        else syz[ sayz[i]/3 + nx ] = name;
    }
    if (name < nxz)
    {
        suffix(syz, sayz, nxz, name);
        for (i = 0; i < nxz; i++) syz[sayz[i]] = i + 1;
    }
    else
    {
        for (i = 0; i < nxz; i++) sayz[syz[i] - 1] = i;
    }
    for (i = j = 0; i < nxz; i++)
    if (sayz[i] < nx) sx[j++] = 3 * sayz[i];
    radix(sx, sax, s, nx, k);
    for (p=0, t=nx-ny, e=0; e < n; e++)
    {
        i = geti(t, nx, sayz); j = sax[p];
        if ( sayz[t] < nx ?leq(s[i], syz[sayz[t]+nx], s[j], syz[j/3]) :
            leq(s[i], s[i+1], syz[sayz[t]-nx+1],
        s[j], s[j+1], syz[j/3+nx]) )
        {
            sa[e] = i;
            if (++t == nxz)
            {
                for (e++; p < nx; p++, e++)
                sa[e] = sax[p];
            }
        }
        else
        {
            sa[e] = j;
            if (++p == nx) for (++e; t < nxz; ++t, ++e)
            sa[e] = geti(t, nx, sayz);
        }
    }
}
void makesa()
{
    memset(buf, 0, 4 * n * sizeof(int));
    memset(sa, 0, 4 * n * sizeof(int));
    for (int i=0; i<n; ++i) buf[i] = s[i] & 0xff;
    suffix(buf, sa, n, 255);
}
void getRank()
{
    for(int i = 1;i < n; ++ i)
        rank[sa[i]] = i;
}
void lcp()
{ // O(4 * N)
    int i, j, k;
    for (j = rank[height[i=k=0]=0]; i < n - 1; i++, k++)
        while (k >= 0 && s[i] != s[ sa[j-1] + k ])
            height[j] = (k--), j = rank[ sa[j] + 1 ];
}
int main()
{
    while(scanf("%s", s) && s[0] != '#')
    {
        n = strlen(s) + 1;
        makesa();
        getRank();
        lcp(); int ans = 0, minid, maxid;
        for(int i = 1; i <= (n >> 1); i++)
        {
            minid = 1200, maxid = -1;
            for(int j = 2; j < n; j++)
                if (height[j] >= i)
                {
                    if (sa[j - 1] < minid) minid = sa[j - 1];
                    if (sa[j - 1] > maxid) maxid = sa[j - 1];
                    if (sa[j] < minid) minid = sa[j];
                    if (sa[j] > maxid) maxid = sa[j];
                }
                else
                {
                    if (maxid != -1 && minid + i <= maxid) ans++;
                    minid = 1200, maxid = -1;
                }
            if (maxid != -1 && minid + i <= maxid) ans++;
        }
        printf("%d\n", ans);
    }
}
 

真弱真窩囊真廢柴。唉。弱爆了。