POJ 3261 (后缀数组 二分) Milk Patterns

这道题和UVa 12206一样，求至少重复出现k次的最长字串。

首先还是二分最长字串的长度len，然后以len为边界对height数组分段，如果有一段包含超过k个后缀则符合要求。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 20000 + 10;
const int maxm = 1000000 + 10;

int s[maxn];
int sa[maxn], height[maxn], rank[maxn];
int t[maxn], t2[maxn], c[maxm];
int n, k;

void build_sa(int m)
{
    int i, *x = t, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1)
    {
        int p = 0;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(i = 1; i < n; i++)
            x[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k] ? p - 1 : p++;
        if(p >= n) break;
        m = p;
    }
}

void build_height()
{
    int i, j, k = 0;
    for(i = 0; i < n; i++) rank[sa[i]] = i;
    for(i = 0; i < n; i++)
    {
        if(k) k--;
        j = sa[rank[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        height[rank[i]] = k;
    }
}

bool ok(int len)
{
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
        if(i == 0 || height[i] < len) cnt = 0;
        if(++cnt >= k) return true;
    }
    return false;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    scanf("%d%d", &n, &k);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &s[i]);
        s[i]++;
    }
    s[n] = 0;
    build_sa(1000002);
    build_height();

    int L = 1, R = n;
    while(L < R)
    {
        int M = (L + R + 1) / 2;
        if(ok(M)) L = M;
        else R = M - 1;
    }

    printf("%d
", L);

    return 0;
}