POJ 3261Milk Patterns
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Description
Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.
To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.
Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.
Input
Lines 2.. N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.
Output
Sample Input
8 212323231
Sample Output
4
后缀数组加二分验证
#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)#define lson x << 1, l, mid#define rson x << 1 | 1, mid + 1, rusing namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-8;const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e6 + 10;int T, m;struct Sa{int s[N];int rk[2][N], sa[N], h[N], w[N], now, n;//int rmq[N][20], lg[N];bool GetS(){if (scanf("%d%d", &n, &m) == EOF) return false;rep(i, 1, n) scanf("%d", &s[i]);return true;}void getsa(int z, int &m){int x = now, y = now ^= 1;rep(i, 1, z) rk[y][i] = n - i + 1;for (int i = 1, j = z; i <= n; i++)if (sa[i] > z) rk[y][++j] = sa[i] - z;rep(i, 1, m) w[i] = 0;rep(i, 1, n) w[rk[x][rk[y][i]]]++;rep(i, 1, m) w[i] += w[i - 1];per(i, n, 1) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];for (int i = m = 1; i <= n; i++){int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;}}void getsa(int m){now = 0;//n = strlen(s + 1);rep(i, 1, m) w[i] = 0;rep(i, 1, n) w[s[i]]++;rep(i, 1, m) rk[1][i] = rk[1][i - 1] + (bool)w[i];rep(i, 1, m) w[i] += w[i - 1];rep(i, 1, n) rk[0][i] = rk[1][s[i]];rep(i, 1, n) sa[w[s[i]]--] = i;for (int x = 1, y = rk[1][m]; x <= n && y < n; x <<= 1) getsa(x, y);for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : 0)while (s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;}/*void getrmq(){lg[1] = 0;rep(i, 2, n) rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;for (int i = 1; (1 << i) <= n; i++){rep(j, 2, n){if (j + (1 << i) > n + 1) break;rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);}}}int lcp(int x, int y){int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);}*/bool check(int x){int y = 1;rep(i, 2, n){if (h[i] >= x) y++;else if (y >= m) return true; else y = 1;}return y >= m;}int work(){int l = 0, r = n - 1;while (l <= r){if (check(l + r >> 1)) l = (l + r >> 1) + 1;else r = (l + r >> 1) - 1;}return r;}}sa;int main(){while (sa.GetS()){sa.getsa(N - 1);printf("%d\n", sa.work());}return 0;}
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