【poj 3261】后缀数组

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先求后缀数组，再二分答案。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <string>#define Rep(i, x, y) for (int i = x; i <= y; i ++)#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)#define Dwn(i, x, y) for (int i = x; i >= y; i --)using namespace std;typedef long long LL;const int N = 20005, M = 1000005;int n, a[N], sa[N], ht[N], c[M], t1[N], t2[N], rk[N], m;bool Cmp(int *r, int x, int y, int l) { return r[x] == r[y] && r[x+l] == r[y+l]; }void Bsa(int *a, int n, int m) {int *x = t1, *y = t2, p = 0;Rep(i, 0, m) c[i] = 0;Rep(i, 0, n) c[x[i] = a[i]] ++;Rep(i, 1, m) c[i] += c[i-1];Rep(i, 0, n) sa[-- c[ x[i] ]] = i;for (int j = 1; p <= n; j <<= 1, m = p) {p = 0;Rep(i, n-j+1, n) y[p ++] = i;Rep(i, 0, n) if (sa[i] >= j) y[p ++] = sa[i] - j;Rep(i, 0, m) c[i] = 0;Rep(i, 0, n) c[ x[y[i]] ] ++;Rep(i, 1, m) c[i] += c[i-1];Dwn(i, n, 0) sa[-- c[ x[y[i]] ]] = y[i];swap(x, y), p = 1, x[ sa[0] ] = 0;Rep(i, 1, n) x[ sa[i] ] = Cmp(y, sa[i], sa[i-1], j) ? p - 1 : p ++;}}void Bh(int *a, int n) {Rep(i, 0, n) rk[ sa[i] ] = i;int k = 0;Rep(i, 0, n-1) {if (k) k --;int j = sa[ rk[i] - 1 ];while (a[i+k] == a[j+k]) k ++;ht[ rk[i] ] = k;}}int main(){scanf ("%d%d", &n, &m), m --;Rep(i, 0, n-1) scanf ("%d", &a[i]), a[i] ++;Bsa(a, n, M-2), Bh(a, n);int l = 0, r = n;while (l < r) {int mid = (l + r + 1) >> 1, o = 0, f = 0;Rep(i, 2, n) {if (ht[i] >= mid) o ++;else o = 0;if (o >= m) { f = 1; break; }}if (f) l = mid;else r = mid - 1;}printf ("%d\n", l);    return 0;}

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