POJ 1743 后缀数组

来源：互联网发布：欧洲难民知乎编辑：程序博客网时间：2024/06/06 18:16

题目的实质就是，不可重叠最长重复字串。

对于音调问题，实际上就是几个数字的 1 2 3 4 和 5 6 7 8 是一样。也就是随便几个数字，f[i] - f[i -1] 和 f[k] - f[k -1] 是一样，就行了。

也就是吧所有数字的差保存起来，求后缀数组，然后二分答案即可。

记住：答案小于5，算无解。（最近在学吉他，这是不是和弦的问题……楼教主懂的真多）

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int max_n = 20000 + 10;int n;int text[max_n];int wa[max_n], wb[max_n], wv[max_n], tub[max_n];int rank[max_n], sa[max_n], height[max_n], a[max_n];inline bool 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)tub[i] = 0;for (i = 0; i != n; ++ i)tub[x[i] = r[i]] ++;for (i = 1; i != m; ++ i)tub[i] += tub[i - 1];for (i = n - 1; i >= 0; -- i)sa[-- tub[x[i]]] = i;for (j = 1, p = 1; p != n; m = p, j *= 2){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)tub[i] = 0;for (i = 0; i != n; ++ i)tub[wv[i]] ++;for (i = 1; i != m; ++ i)tub[i] += tub[i - 1];for (i = n - 1; i >= 0; -- i)sa[-- tub[wv[i]]] = y[i];for (t =x, x=y,y=t, p = 1, i = 1, x[sa[0]] = 0; i != n; ++ i)x[sa[i]] = cmp(y, sa[i], sa[i - 1], j) ? p - 1: p ++;}}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);}bool check(int k)//长度为k的是否存在{int max_wz = 1, min_wz = 1;for (int i = 2; i <= n; ++ i){if (height[i] < k){if (max_wz - min_wz >= k)return true;max_wz = min_wz = sa[i];continue;}max_wz = max(max_wz, sa[i]);min_wz = min(min_wz, sa[i]);}if (max_wz - min_wz >= k)return true;return false;}void doit(){int L = 0, R = n;while (L + 1 < R){int mid = (L + R)/2;if (check(mid))L = mid;else R = mid;}if (L + 1 < 5)cout<<0<<endl;else cout<<L + 1<<endl;}int main(){ios::sync_with_stdio(false);while (1){cin >> n;if (n == 0)break;for (int i = 0; i != n; ++ i)cin >> a[i];text[0] = 100;for (int i = 1; i != n; ++ i)text[i] = a[i] -  a[i - 1] + 100; //做了差但是+100用以保障都是正数text[n] = 0;da(text, sa, n + 1, 200);calheight(text, sa, n);doit();}return 0;}

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