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Musical Theme
Time Limit:1000MSMemory Limit:30000K
Total Submissions:19985Accepted:6835
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
3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800
Sample Output
5
Hint
Use scanf instead of cin to reduce the read time.
题意：给出n个范围在1~88的音符，求最长的重复主题的长度
重复主题：1，不可重叠； 2，某子串所有元素加减某个数可得到另一个子串；
如：1 4 3 6 24 5 6 34 7 6 9 中， 4 3 6 和 7 6 9 就是重复的主题，因为 4 3 6 中每个元素加上3 就跟 跟它主题重复的子串相同。
思路：令所有 val[i] = val[i+1] – val[i]，，则不需要进行转换，然后每个元素都加上88，再在末尾添加个元素0，之后就是范围在0~175的后缀数组求不可重叠的最长重复子串。最终结果+1.
留意一下这个例子：
11
1 2 3 4 5 6 7 8 9 10 11
答案应该是5， 我相信比较多人会输出6
虽然POJ无视此问题，但我还是在输出时做了一点小判断。。。
#include <stdio.h> #include <stdlib.h> #include <string.h> #define MS(x, y) memset(x, y, sizeof(x)) const int MAXN = 20000+10; const int INF = 1<<30;int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN]; int rank[MAXN],r[MAXN],sa[MAXN],height[MAXN]; int num[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 i, j, p, *x = wa, *y = wb, *t;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; }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-1; height[rank[i++]] = k)for(k ? k– : 0,j=sa[rank[i]-1]; r[i+k] == r[j+k]; k++);return; } int main(){//freopen("in.txt", "r", stdin);int n;while(~scanf("%d", &n), n){int i;for(i=0; i<n; i++)scanf("%d",r + i);if(n < 10){printf("0\n");continue;}for(i=0; i<n-1; i++)r[i] = r[i+1] – r[i];for(i=0; i<n-1; i++)r[i] += 88;r[n-1] = 0;da(r, sa, n, 200);calheight(r, sa, n);int left = 1, right = n;int ok;int ans = 0;int beg, end;while(left <= right){int mid = left + (right – left)/2;//printf("%d\n", mid);beg = end = ok = 0;for(int i=2; i<n; i++){if(height[i] >= mid){//确定某一组的起点终点if(!beg) beg = i;end = i;}if((beg && end) && (i == n – 1 || height[i] < mid)){int minn = INF, maxn = -1;for(int i=beg-1; i<=end; i++){if(sa[i] > maxn) maxn = sa[i];if(sa[i] < minn) minn = sa[i];}if(maxn – minn >= mid){//子串不重叠ok = 1;break;}beg = end = 0;}}if(ok) ans = mid;if(ok) left = mid + 1;else right = mid – 1;}if(ans < 4) printf("0\n");else{ans += 1;if(ans-1 == n/2 && n%2) ans -= 1;printf("%d\n", ans);}}return 0;}
人生就像一场旅行，不必在乎目的地，
