Power Strings POJ - 2406

Power Strings

 POJ - 2406 

ｋｍｐ可以过的

学了ｓａ就用ｓａ搞一下，，一直ＴＬＥ．．．

好像要用另外一种方法实现，还没学=_=

先放上超时的代码吧

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000010;
char s[maxn];
int sa[maxn], t1[maxn], t2[maxn], c[maxn];
int h[maxn], rk[maxn];
int n;
void build_sa(int n, int m){
    int i, *x = t1, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i-1];
    for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1){
        int p = 0;
        for(i = n-k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k; 
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i< m; i++) c[i] += c[i-1];
        for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(i = 1; i < n; i++) 
          x[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]? p-1 : p++;
        if(p>=n) break;
        m=p;
    }
}
void geth(int n){
    int i, j, k=0;
    for(i = 0; i < n; i++) rk[sa[i]]=i; 
    for(i = 0; i < n; i++) {
        if(k) k--;
        j = sa[rk[i]-1];
        while(s[i+k]==s[j+k]) k++;
        h[rk[i]] = k;
    }
}
int main(){
    while(scanf("%s", s)){
        if(s[0] == '.') break;
        int n = strlen(s) + 1;  //!!!
        build_sa(n, 255);
        geth(n);
        int k;
        n--;
        for(k = 1; k <= n; k++){
            if(n%k==0 && h[rk[0]] == n-k) break;
        }
        printf("%d\n", n/k);
    }
}

 

DC3

找了个模板，，过了，日后有空再细看～

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
/*
 * 后缀数组
 * DC3算法,复杂度O(n)
 * 所有的相关数组都要开三倍 
 */
const int MAXN = 1000010;
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x)-tb) * 3 + 2)

int wa[MAXN * 3], wb[MAXN * 3], wv[MAXN * 3], wss[MAXN * 3];

int c0(int *r, int a, int b)
{
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}

int c12(int k, int *r, int a, int b)
{
    if(k == 2)
    {
        return r[a] < r[b] || (r[a] == r[b] && c12(1, r, a + 1, b + 1));
    }
    else
    {
        return r[a] < r[b] || (r[a] == r[b] && wv[a + 1] < wv[b + 1]);
    }
}

void sort(int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i < n; i++)
    {
        wv[i] = r[a[i]];
    }
    for (i = 0; i < m; i++)
    {
        wss[i] = 0;
    }
    for (i = 0; i < n; i++)
    {
        wss[wv[i]]++;
    }
    for (i = 1; i < m; i++)
    {
        wss[i] += wss[i - 1];
    }
    for (i = n - 1; i >= 0; i--)
    {
        b[--wss[wv[i]]] = a[i];
    }
}

void dc3(int *r, int *sa, int n, int m)
{
    int i, j, *rn = r + n;
    int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p;
    r[n] = r[n+1] = 0;
    for (i = 0; i < n; i++)
    {
        if (i % 3 != 0)
        {
            wa[tbc++] = i;
        }
    }
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
    {
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
    }
    if (p < tbc)
    {
        dc3(rn, san, tbc, p);
    }
    else
    {
        for (i = 0; i < tbc; i++)
        {
            san[rn[i]] = i;
        }
    }
    for (i = 0; i < tbc; i++)
    {
        if (san[i] < tb)
        {
            wb[ta++] = san[i] * 3;
        }
    }
    if (n % 3 == 1)
    {
        wb[ta++] = n - 1;
    }
    sort(r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++)
    {
        wv[wb[i] = G(san[i])] = i;
    }
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
    {
        sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    }
    for (; i < ta; p++)
    {
        sa[p] = wa[i++];
    }
    for (; j < tbc; p++)
    {
        sa[p] = wb[j++];
    }
}

//  str和sa也要三倍
void da(int str[], int sa[], int rk[], int h[], int n, int m)
{
    for (int i = n; i < n * 3; i++)
    {
        str[i] = 0;
    }
    dc3(str, sa, n+1, m);
    int i, j, k = 0;
    for (i = 0; i <= n; i++)
    {
        rk[sa[i]] = i;
    }
    for (i = 0; i < n; i++)
    {
        if(k)
        {
            k--;
        }
        j = sa[rk[i] - 1];
        while (str[i + k] == str[j + k])
        {
            k++;
        }
        h[rk[i]] = k;
    }
}
//sa[i] ：排名第 i 的后缀在哪（i 从 1 开始）
//rk[i]：后缀 i 排第几 （i 从 0 开始）
//h[i]：排名为 i 和 i-1 的两个后缀的最长公共前缀(LCP)长度 （i 从 2 开始）
int s[MAXN * 3];
char str[MAXN * 3];
int sa[MAXN * 3];
int h[MAXN * 3],rk[MAXN * 3];

int main(){
    //freopen("in.txt", "r", stdin);
    while(scanf("%s", str)){
        if(str[0] == '.') break;
        int n = strlen(str);
        for(int i = 0; i <  n; i++) s[i] = str[i] - 'a' + 1;
        da(s, sa, rk, h, n, 300);
        int k;
        for(k = 1; k <= n; k++){
            if(n%k==0 && rk[0] == rk[k]+1 && h[rk[0]] == n-k) break;
        }
        if(k == n+1) k--;
        printf("%d\n", n/k);
    }
}

 

 

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高级数据结构p399习题1