HDU 5219 Repeating 后缀数组 + 莫比乌斯函数

来源：互联网发布：爬虫java 教程编辑：程序博客网时间：2024/05/19 18:14

题目大意：

就是现在给出一个长度不超过100100的只包含小写字母的字符串, 求问这个串有多少个字串没有循环节, 如abab有循环节ab, 而aba, a, abc没有

大致思路：

好久没写题解了...补一个历史遗留的坑...

首先这题要先枚举循环节长度, 然后找到所有的可以以这个长度为循环节的串

根据每次枚举的循环节长度将串分成多个长度为L的组, 然后用后缀数组找出连续的一整段子串, 这个子串的任意长度是L的n(n >= 2)倍长度的子串都是不合法串, 那么因为这样枚举的时候会出现不同的L枚举到同一个串的情况, 这里用莫比乌斯函数来处理一下就好了

例如对于某个子串 S = abababababab当枚举循环节长度为2的时候S作为6个循环节计算一次, 枚举循环节长度为2, 3的时候也会算到, 循环节个数依次是3, 2, 那么答案就根据莫比乌斯函数来累加,, 为mu[6] + mu[3] + mu[2] = -1也就只减去了一次, 同理S = aaaaaaaaaaaa在计算的时候循环节个数在枚举循环节长度为1, 2, 3, 4, 6的时候算到, 计数就是mu[12] + mu[6] + mu[4] + mu[3] + mu[2] = -1, 想到莫比乌斯函数就是一个简单的计数了..

代码如下：

Result : Accepted Memory : 22052 KB Time : 1263 ms

/* * Author: Gatevin * Created Time:  2015/9/26 18:55:50 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define ws Ws#define rank rrrankconst int maxn = 100200 + 1;//Mobius筛到maxn - 1int prime[maxn], vis[maxn], mu[maxn];int tot;void Mobius(){    mu[1] = 1;    for( int i = 2 ; i < maxn ; i ++ ){        if( !vis[i] )            prime[++tot] = i, mu[i] = -1;        for( int j = 1 ; j <= tot && i * prime[j] < maxn ; j ++ ){            vis[i*prime[j]] = 1;            if( i % prime[j] == 0 ){                mu[i*prime[j]] = 0;                break;            }            else                mu[i*prime[j]] = - mu[i];        }    }    return;}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 *x = wa, *y = wb, *t, i, j, p;    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 <<= 1, 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 rank2[maxn], height2[maxn];void calheight2(int *r, int *sa, int n){    int i, j, k = 0;    for(i = 1; i <= n; i++) rank2[sa[i]] = i;    for(i = 0; i < n; height2[rank2[i++]] = k)        for(k ? k-- : 0, j = sa[rank2[i] - 1]; r[i + k] == r[j + k]; k++);    return;}int dp[maxn][20];void initRMQ(int n){    for(int i = 1; i <= n; i++) dp[i][0] = height[i];    for(int j = 1; (1 << j) <= n; j++)        for(int i = 1; i + (1 << j) - 1 <= n; i++)            dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);    return;}int dp2[maxn][20];void initRMQ2(int n){    for(int i = 1; i <= n; i++) dp2[i][0] = height2[i];    for(int j = 1; (1 << j) <= n; j++)        for(int i = 1; i + (1 << j) - 1 <= n; i++)            dp2[i][j] = min(dp2[i][j - 1], dp2[i + (1 << (j - 1))][j - 1]);    return;}int askRMQ(int a, int b){    int ra = rank[a], rb = rank[b];    if(ra > rb) swap(ra, rb);    int k = 0;    while((1 << (k + 1)) <= rb - ra) k++;    return min(dp[ra + 1][k], dp[rb - (1 << k) + 1][k]);}int len;int askRMQ2(int a, int b){    a = len - a - 1;    b = len - b - 1;    int ra = rank2[a], rb = rank2[b];    if(ra > rb) swap(ra, rb);    int k = 0;    while((1 << (k + 1)) <= rb - ra) k++;    return min(dp2[ra + 1][k], dp2[rb - (1 << k) + 1][k]);}char in[maxn];int s[maxn], sa[maxn];int main(){    Mobius();    int T;    scanf("%d", &T);    while(T--)    {        scanf("%s", in);        len = strlen(in);        for(int i = 0; i < len; i++)            s[i] = in[i] - 'a' + 1;        s[len] = 0;        da(s, sa, len + 1, 28);        calheight(s, sa, len);        initRMQ(len);                for(int i = 0; i < len; i++)            s[i] = in[len - 1 - i] - 'a' + 1;        s[len] = 0;        da(s, sa, len + 1, 28);        calheight2(s, sa, len);        initRMQ2(len);                        lint ans = 0;        for(int L = 1; L <= (len >> 1); L++)//枚举循环节长度        {            int n = len;            int blocks = n / L + (n % L != 0);            int now = 1;            while(now < blocks)            {                if(now + 1 < blocks)                {                    int len2 = askRMQ2((now + 1)*L - 1, now*L - 1);//用后缀数组向前和向后找到区间[L, R]这个串中任意长度是L的2以上倍数的子串都不合法                    int len1 = askRMQ(now*L, (now + 1)*L);                    int totlen = L + len2 + len1;                    int cnt = totlen / L;                    for(int i = 2; i <= cnt; i++)                    {                        ans += mu[i]*(totlen - L*i + 1);                    }                    //now = now + (len1 / L) + (len1 % L != 0) + 1;注意不要跳掉len1 % L != 0的时候中间那个没用完的段                    //printf("find1 L = %d, ans = %I64d, totlen = %d, now = %d\n", L, ans, totlen, now);                    now = now + (len1 / L) + 1;                                    }                else                {                    if(n % L != 0) break;//注意末尾...刚开始这里没注意到...对于abababababab一个个数终于发现了这个错误                    int totlen = (len - now*L) + askRMQ2(len - 1, now*L - 1);                    int cnt = totlen / L;                    //int num = totlen % L + 1;                    for(int i = 2; i <= cnt; i++)                        ans += mu[i]*(totlen - L*i + 1);                    //printf("find2 L = %d, ans = %I64d, totlen = %d, now = %d\n", L, ans, totlen, now);                    now = blocks;                }            }        }        printf("%I64d\n", (lint)len*(len + 1LL) / 2LL + ans);    }    return 0;}/*5aaccaaabcabcaaaaaaabababababab1820653*/

0 0