poj 3261 Milk Patterns(最长至少k次重复子串,后缀数组基础题)

题目:http://poj.org/problem?id=3261

题目大意:给你一个字符串,让你输出它的最长的至少k次重复的子串的长度。

思路:先离散一下,然后构造后缀数组,然后二分答案,每次一边扫 height 就行了。

代码如下:

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

const int MAXN = 22222;

int s[MAXN];

int sa[MAXN],rank[MAXN],height[MAXN];

int t[MAXN],t2[MAXN],c[MAXN];

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

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

bool flag[MAXN];

int check(int mid,int n,int k)
{
    int cc = 0;
    memset(flag,0,sizeof(flag));
    for(int i = 2;i <= n;i++)
    {
        //printf("i = %d,hi = %d\n",i,height[i]);
        if(height[i] >= mid)
        {
            if(flag[i-1] == 0)
            {
                flag[i-1] = 1;
                cc++;
            }
            if(flag[i] == 0)
            {
                flag[i] = 1;
                cc++;
            }
        }
        else
        {
            cc = 0;
            memset(flag,0,sizeof(flag));
        }

        if(cc >= k) return 1;
    }

    return 0;
}

int slove(int n,int k)
{
    int l = 1,r = n;
    int ans = 0;
    while(l <= r)
    {
        int mid = (l+r) >> 1;
        //printf("l = %d,r = %d,mid = %d\n",l,r,mid);
        if(check(mid,n,k))
        {
            ans = mid;
            l = mid+1;
        }
        else r = mid - 1;
    }
    return ans;
}

map <int,int> mm;

int num[MAXN];

int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        mm.clear();
        int tot = 1;
        for(int i = 0;i < n;i++)
        {
            scanf("%d",&num[i]);
            if(!mm[num[i]])
                mm[num[i]] = tot++;
        }
        for(int i = 0;i < n;i++)
            s[i] = mm[num[i]];
        s[n] = 0;
        //for(int i = 0;i <= n;i++)
            //printf("i = %d,si = %d\n",i,s[i]);
        da(n+1,tot);
        printf("%d\n",slove(n,k));
    }
    return 0;
}


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