Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings. Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it:
is at least five notes longappears (potentially transposed — see below) again somewhere else in the piece of musicis disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)
Transposed means that a constant positive or negative value is added to every note value in the theme subsequence. Given a melody, compute the length (number of notes) of the longest theme. One second time limit for this problem’s solutions!
Input The input contains several test cases. The first line of each test case contains the integer N. The following n integers represent the sequence of notes. The last test case is followed by one zero.
Output For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.
Sample Input
30 25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18 82 78 74 70 66 67 64 60 65 80 0
Sample Output
5
Hint Use scanf instead of cin to reduce the read time.
Source LouTiancheng@POJ
求最长不可重叠子串，，可以后缀数组+二分解决 先把输入的数字前后两两做差，然后建立后缀数组，二分即可
/*************************************************************************> File Name: POJ1743.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年03月31日 星期二 15时43分29秒 ************************************************************************/;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;LL;typedef pair <int, int> PLL;class SuffixArray{public:N = 20010;int init[N];int X[N];int Y[N];int Rank[N];int sa[N];int height[N];int buc[N];int size;void clear(){size = 0;}void insert(int n){init[size++] = n;}bool cmp(int *r, int a, int b, int l){return (r[a] == r[b] && r[a + l] == r[b + l]);}void getsa(int m = 256){init[size] = 0;int l, p, *x = X, *y = Y, n = size + 1;for (int i = 0; i < m; ++i){buc[i] = 0;}for (int i = 0; i < n; ++i){buc[x[i] = init[i]]++;}for (int i = 1; i < m; ++i){buc[i] += buc[i – 1];}for (int i = n – 1; i >= 0; –i){sa[–buc[x[i]]] = i;}for (l = 1, p = 1; l <= n; m = p, l *= 2){p = 0;for (int i = n – l; i < n; ++i){y[p++] = i;}for (int i = 0; i < n; ++i){if (sa[i] >= l){y[p++] = sa[i] – l;}}for (int i = 0; i < m; ++i){buc[i] = 0;}for (int i = 0; i < n; ++i){++buc[x[y[i]]];}for (int i = 1; i < m; ++i){buc[i] += buc[i – 1];}for (int i = n – 1; i >= 0; –i){sa[–buc[x[y[i]]]] = y[i];}int i;for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i){x[sa[i]] = cmp(y, sa[i – 1], sa[i], l) ? p – 1 : p++;}if (p >= n){break;}}}void getheight(){int h = 0;for (int i = 0; i <= size; ++i){Rank[sa[i]] = i;}height[0] = 0;for (int i = 0; i < size; ++i){if (h > 0){–h;}int j = sa[Rank[i] – 1];for (; i + h < size && j + h < size && init[i + h] == init[j + h]; ++h);height[Rank[i] – 1] = h;}}bool judge(int k){int maxs = sa[1], mins = sa[1];for (int i = 1; i < size; ++i){if (height[i] < k){maxs = mins = sa[i + 1];}else{maxs = max(maxs, sa[i + 1]);mins = min(mins, sa[i + 1]);if (maxs – mins > k){return 1;}}}return 0;}void solve(){int l = 1, r = size;int mid;int ans = 0;while (l <= r){int mid = (l + r) >> 1;if (judge(mid)){l = mid + 1;ans = mid;}else{r = mid – 1;}}++ans;printf(“%d\n”, ans >= 5 ? ans : 0);}}SA;int val[20010];int main(){int n;while (~scanf(“%d”, &n), n){SA.clear();for (int i = 1; i <= n; ++i){scanf(“%d”, &val[i]);}for (int i = n; i >= 2; –i){val[i] = val[i] – val[i – 1] + 90;}for (int i = 2; i <= n; ++i){SA.insert(val[i]);}SA.getsa();SA.getheight();SA.solve();}return 0;}
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