POJ3261 Milk Patterns

Milk Patterns

题目大意

求一个数串(数集为0~10000000)中最长至少重复k次的子串。

题解

SA二分分组。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <iostream>
void swap(int &a, int &b){int tmp = a;a = b, b = tmp;}
void swap(int* &a, int* &b){int* tmp = a;a = b;b = tmp;}
int max(int a, int b){return a > b ? a : b;}
int min(int a, int b){return a < b ? a : b;}
int lowbit(int x){return x & (-x);}
void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 20000 + 10; 
const int MAXNUM = 1000000 + 10;


struct SuffixArray
{
    int s[MAXN], sa[MAXN], rank[MAXN], height[MAXN];
    int t[MAXNUM], t2[MAXNUM], c[MAXNUM];
    int n;
    void clear()
    {
        n = 0;
        memset(sa, 0, sizeof(sa));
    }
    
    void build_sa(int m)
    {
        ++ n;
        int i,*x = t,*y = t2;
        for(i = 0;i < m;++ i) c[i] = 0;
        for(i = 0;i < n;++ i) x[i] = s[i];
        for(i = 0;i < n;++ i) ++ c[x[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[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;
        }
        -- n;
    }
    
    void build_height()
    {
        int i, j, k = 0;
        for(i = 1;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;
        }
    }
}A;

int n, k;
int l[MAXN], r[MAXN], tot;

bool check(int x)
{
    l[tot = 1] = 1;
    for(int i = 2;i <= n;++ i)
    {
        while(A.height[i] >= x) ++ i;
        r[tot] = i - 1;
        l[++ tot] = i;
    }
    r[tot] = n;

    for(int i = 1;i <= tot;++ i)
        if(r[i] - l[i] + 1 >= k)
            return 1;
    return 0;
}

int main()
{
    read(n), read(k);
    for(int i = 0;i < n;++ i) read(A.s[i]);
    A.s[n] = 0;
    A.n = n;
    A.build_sa(MAXNUM);
    A.build_height();
    int l = 0, r = n, mid, ans = 0;
    while(l <= r)
    {
        mid = (l + r) >> 1;
        if(check(mid)) ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    printf("%d", ans);
    return 0;
} 
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