SPOJ 687 REPEATS Repeats 后缀数组 + RMQ预处理

题目大意：

对于给出的字符串(长度<= 50000,只包含字符'a'或'b')找到最大的k使得存在某个字符串t重复k次是给出的字符串的子串

大致思路：

就是POJ 3693那题的简单版本....不需要找到字典序最小的....

首先有这样一个事实: 对于任何一个子串, repetition number >= 1, 所以对于repetition number为1的只需要找到字典序最小的那个字母即可, 那么我们只考虑repetition number >= 2的情况, 如果每一个循环节的长度为len, 那么在原字符串S中, S[i*len]与S[(i + 1)*len]一定会被包含在答案的子串当中那么枚举可能的答案的循环节的长度, 然后枚举可能的位置, 对于每一组可能被包含的位置S[i*len], S[(i + 1)*len]求出其对应后缀的最长公共前缀长度L, 则该循环节至少循环了L/len + 1次, 但是当L%len != 0时, 后面多余出来的部分(长度L%len的部分)可能和前面的拼凑成循环节, 于是对于位置i*len - (len - L % len)和(i + 1)*len - (len - L % len)求其后缀的最长公共前缀长度, 如果大于之前的结果,自然就说明从这个位置开始可以比之前多一个循环节, 于是这样枚举得到最多循环次数

 * Author: Gatevin * Created Time:  2015/2/9 15:21:08 * File Name: Iris_Freyja.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 maxn 50233int 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 *= 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 s[maxn], sa[maxn];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 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 main(){    int H;    scanf("%d", &H);    int n;    while(H-->0)    {        scanf("%d", &n);        char tmp;        for(int i = 0; i < n; i++)        {            getchar();            scanf("%c", &tmp);            s[i] = tmp - 'a' + 1;        }        s[n] = 0;        da(s, sa, n + 1, 3);        calheight(s, sa, n);        initRMQ(n);        int anstimes = 0;        for(int len = 1; len <= n; len++)            for(int i = 0; i + len < n; i += len)            {                int lcp = askRMQ(i, i + len);                int times = lcp / len + 1;                int k = i - (len - lcp % len);                if(k >= 0 && lcp % len)                    if(askRMQ(k, k + len) >= lcp)                        times++;                if(times > anstimes)                    anstimes = times;            }        printf("%d\n", anstimes);    }    return 0;}</span>