Milk Patterns 【POJ - 3261】【后缀数组】

题目链接

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  题意：给一个长度为N的串，问你其中出现次数大于等于N次的子串的最长长度。

  思路：直接上后缀数组，但是要有个注意事项，也就是值是从0～1e6的，所以呢，SA处理的时候，需要考虑“0”的情况。那么，就是给每个元素“+1”处理，就可以做到这个了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e6 + 7;
struct SA
{
    int n, m;
    int s[maxN];
    int y[maxN], x[maxN], c[maxN], sa[maxN], rk[maxN], height[maxN];
    inline void get_SA()
    {
        for(int i=1; i<=m; i++) c[i] = 0;   //桶的初始化
        for(int i=1; i<=n; i++) ++c[x[i] = s[i]];
        for(int i=2; i<=m; i++) c[i] += c[i - 1];   //利用差分前缀和的思想知道每个关键字最多是在第几名
        for(int i=n; i>=1; i--) sa[c[x[i]]--] = i;
        for(int k=1; k<=n; k<<=1)
        {
            int num = 0;
            for(int i=n - k + 1; i<=n; i++) y[++num] = i;
            for(int i=1; i<=n; i++) if(sa[i] > k) y[++num] = sa[i] - k; //是否可以作为第二关键字
            for(int i=1; i<=m; i++) c[i] = 0;
            for(int i=1; i<=n; i++) c[x[i]]++;  //因为上一次循环已经求出这次的第一关键字了
            for(int i=2; i<=m; i++) c[i] += c[i - 1];
            for(int i=n; i>=1; i--) //在同一第一关键字下，按第二关键字来排
            {
                sa[c[x[y[i]]]--] = y[i];
                y[i] = 0;
            }
            swap(x, y);
            x[sa[1]] = 1; num = 1;
            for(int i=2; i<=n; i++)
            {
                x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? num : ++num;
            }
            if(num == n) break;
            m = num;
        }
    }
    inline void get_height()
    {
        int k = 0;
        for(int i=1; i<=n; i++) rk[sa[i]] = i;
        for(int i=1; i<=n; i++)
        {
            if(rk[i] == 1) continue;    //第一名的height为0
            if(k) k--;  //height[i] >= height[i - 1] - 1
            int j = sa[rk[i] - 1];
            while(j + k <= n && i + k <= n && s[i + k] == s[j + k]) k++;
            height[rk[i]] = k;
        }
    }
    inline void clear()
    {
        n = 0; m = 1e6 + 2;
    }
} sa;
struct BIT_Tree
{
    int tree[maxN << 2];
    inline void buildTree(int rt, int l, int r)
    {
        if(l == r) { tree[rt] = sa.height[l]; return; }
        int mid = HalF;
        buildTree(Lson); buildTree(Rson);
        tree[rt] = min(tree[lsn], tree[rsn]);
    }
    inline int query(int rt, int l, int r, int ql, int qr)
    {
        if(ql <= l && qr >= r) return tree[rt];
        int mid = HalF;
        if(qr <= mid) return query(QL);
        else if(ql > mid) return query(QR);
        else return min(query(QL), query(QR));
    }
} tree;
int N, K, len, s[maxN], ans_len;
int main()
{
    sa.clear();
    scanf("%d%d", &N, &K);
    for(int i=1; i<=N; i++) scanf("%d", &s[i]);
    for(int i=1; i<=N; i++) sa.s[++sa.n] = s[i] + 1;
    sa.s[++sa.n] = 1e6 + 2;
    sa.get_SA();
    sa.get_height();
    tree.buildTree(1, 2, sa.n - 1);
    int l = 1, r = K + 1;
    ans_len = tree.query(1, 2, sa.n - 1, l + 1, r - 1);
    while(r <= N)
    {
        l++; r++;
        ans_len = max(ans_len, tree.query(1, 2, sa.n - 1, l + 1, r - 1));
    }
    printf("%d\n", ans_len);
    return 0;
}