Milk Patterns POJ - 3261（后缀数组+二分）

题意：

　　求可重叠的最长重复子串，但有一个限制条件。。要至少重复k次

解析：

　　二分枚举k，对于连续的height 如果height[i] >= k 说明它们至少有k个元素是重复的，所以判断一下就好了

数据很水

　　输入数据可能为0，所以s[i]++  s[n++] = 0;

要后缀数组要保证末尾加的字符比前面的都小

这是百度百科上的。。这里的r数组即为s数组

 

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 50010, INF = 0x7fffffff;

int a[maxn], s[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn], n;
int ran[maxn], height[maxn];

void get_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 = 0; 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-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    for(i = 0; i < n; i++) ran[sa[i]] = i;
    for(i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[ran[i]-1];
        while(s[i+k] == s[j+k]) k++;
        height[ran[i]] = k;
    }
}
int len;

bool solve(int k)
{
    int cnt = 1;
    for(int i=1; i<n; i++)
        if(height[i] >= k)
        {
            cnt++;
            if(cnt>=len) return true;
        }
        else
            cnt = 1;
    return false;
}

int main()
{
    rd(n); rd(len);
    rep(i, 0, n){
        rd(s[i]);
        s[i]++;
    }
    s[n++] = 0;
    get_sa(20010);
    int l = 0, r = n;
    while(l <= r)
    {
        int mid = l + (r - l) / 2;
        if(solve(mid)) l = mid + 1;
        else r = mid - 1;
    }
    cout<< r <<endl;

    return 0;
}