Longest Repeated Sequence

描述 
You are given a sequence of integers, A = a1, a2, ... an. A consecutive subsequence of A (say ai, ai+1 ... aj) is called a "repeated sequence" if it appears more than once in A (there exists some positive k that ai+k = ai, ai+k+1 = ai+1, ... aj+k = aj) and its appearances are not intersected (i + k > j).

Can you find the longest repeated sequence in A?

输入 
Line 1: n (1 <= n <= 300), the length of A. 
Line 2: the sequence, a1 a2 ... an (0 <= ai <= 100).

输出 
The length of the longest repeated sequence.

样例输入 
5 
2 3 2 3 2

样例输出 
2

不知道这题有没有哪位大侠直接暴力过的，感觉n的级别，三次方的话1s有点悬。。。我的想法是，用二分法确定序列最长的长度。然后为了快速判断某一个长度是否成立，建立一个next数组，保存的是从右边开始跟它相等的最近的一个元素的位置，若右方没有与之相等的元素，next数组值为-1，这样在寻找序列起始位置的时候就可以根据next数组直接往下跳了，节省了很多不必要的判断时间。不过这题我当时做的时候考虑不周，在使用next数组往后跳的时候只选择了第一个满足条件的位置，忘了后面还可能有很多个满足条件的位置，只得了90分，结果是WA。当时只剩下十几分钟，就没来得及细想，后来同学问我是怎么做的时候，我才发现这个问题，可惜已经无法提交了。先把比赛时候的代码贴出来吧。

1 #include <cstdio> 2 #include <cstring> 3  4 const int N = 305; 5  6 int n; 7 int seq[N], next[N]; 8  9 // 判断当前长度是否满足条件10 bool check(int num)11 {12     if (0 == num)13         return true;14     int i, j, k, res, up = n - (num << 1) + 1;15     for (i = 1; i <= up; i++) {16         // next数组为-1，后面没有相等元素，跳过17         if (-1 == next[i])18             continue;19         j = next[i];20         k = j - i;21         // 序列长度至少为num，才开始比较22         while (k < num && -1 != next[j]) {23             k += next[j] - j;24             j = next[j];25         }26         if (k < num)27             continue;28         // 考虑所有满足条件的位置j29         do {30             int l = i, r = j;31             res = 0;32             while (seq[l] == seq[r] && r <= n) {33                 res++;34                 l++, r++;35                 if (res >= num)36                     return true;37             }38             j = next[j];39         } while (-1 != j);40     }41     return false;42 }43 44 int main(void)45 {46     int i, j;47     int left, right, mid;48     while (EOF != scanf("%d", &n)) {49         memset(next, -1, sizeof(next));50         for (i = 1; i <= n; i++) {51             scanf("%d", &seq[i]);52             // 设置元素的next数组，右边与其最接近的相等元素的位置53             for (j = i - 1; j >= 1; j--) {54                 if (seq[j] == seq[i]) {55                     next[j] = i;56                     break;57                 }58             }59         }60         left = 0, right = (n >> 1) + 1;61         // 二分法确定满足条件的最长长度62         while (left <= right) {63             mid = (left + right) >> 1;64             if (check(mid))65                 left = mid + 1;66             else67                 right = mid - 1;68         }69         printf("%d\n", left - 1);70     }71     return 0;72 }