hdu6194 string string string

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6194

题目：

string string string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 486    Accepted Submission(s): 125

Problem Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

 

 

Input

The first line contains an integer T (T≤100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.

 

 

Output

For each test case, print the number of the important substrings in a line.

 

 

Sample Input

2 2 abcabc 3 abcabcabcabc

 

 

Sample Output

6 9

 

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 　　

思路：

　　sam会卡指针版的代码，不明白卡这么点内存有什么意思。

　　后缀自动机做法：建立sam后，统计出cnt数组，这一步要拓扑排序后倒着dp。

　　后缀数组做法：用f[i][j]表示height(i,j)的区间最小值，然后滑动大小为k的窗口，则贡献是max(0,f[i][j]-max(height[i-1],height[j+1]) )

后缀自动机做法：

#include <bits/stdc++.h>

using namespace std;

char ss[100004];
int ans;

struct SAM
{
    static const int MAXN = 100001<<1;//大小为字符串长度两倍
    static const int LetterSize = 26;

    int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];
    int sum[MAXN], tp[MAXN], cnt[MAXN]; //sum,tp用于拓扑排序，tp为排序后的数组

    void init( void)
    {
        last = tot = 1;
        len[1] = 0;
        memset(ch,0,sizeof ch);
        memset(fa,0,sizeof fa);
        memset(cnt,0,sizeof cnt);
    }

    void add( int x)
    {
        int p = last, np = last = ++tot;
        len[np] = len[p] + 1, cnt[last] = 1;
        while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
        if( p == 0)
            fa[np] = 1;
        else
        {
            int q = ch[p][x];
            if( len[q] == len[p] + 1)
                fa[np] = q;
            else
            {
                int nq = ++tot;
                memcpy( ch[nq], ch[q], sizeof ch[q]);
                len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
                while( p && ch[p][x] == q)  ch[p][x] = nq, p = fa[p];
            }
        }
    }

    void toposort( void)
    {
        for(int i = 1; i <= len[last]; i++)   sum[i] = 0;
        for(int i = 1; i <= tot; i++)   sum[len[i]]++;
        for(int i = 1; i <= len[last]; i++)   sum[i] += sum[i-1];
        for(int i = 1; i <= tot; i++)   tp[sum[len[i]]--] = i;
    }
} sam;


int main(void)
{
    //freopen("in.acm","r",stdin);
    int t,k;cin>>t;
    while(t--)
    {
        ans=0;
        sam.init();
        scanf("%d%s",&k,ss);
        for(int i=0,len=strlen(ss);i<len;i++)  sam.add(ss[i]-'a');
        sam.toposort();
        for(int i=sam.tot;i;i--)
        {
            int p=sam.tp[i],fp=sam.fa[p];
            sam.cnt[fp]+=sam.cnt[p];
            if(sam.cnt[p]==k)
                ans+=sam.len[p]-sam.len[fp];
        }
        printf("%d\n",ans);
    }

    return 0;
}

后缀数组做法：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include <iostream>
const int N = 200005;
int sa[N],s[N],wa[N], wb[N], ws[N], wv[N];
int rank[N], height[N];
char ss[N];
bool cmp(int r[], int a, int b, int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void da(int r[], int sa[], int n, int m)
{
    int i, j, p, *x = wa, *y = wb;
    for (i = 0; i < m; ++i) ws[i] = 0;
    for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
    for (i = 1; i < m; ++i) ws[i] += ws[i-1];
    for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, m = p)
    {
        for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
        for (i = 0; i < n; ++i) wv[i] = x[y[i]];
        for (i = 0; i < m; ++i) ws[i] = 0;
        for (i = 0; i < n; ++i) ws[wv[i]]++;
        for (i = 1; i < m; ++i) ws[i] += ws[i-1];
        for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
        for (std::swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
    }
}

void calheight(int r[], int sa[], int n)
{
    int i, j, k = 0;
    for (i = 1; i <= n; ++i) rank[sa[i]] = i;
    for (i = 0; i < n; height[rank[i++]] = k)
        for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
}
struct RMQ
{
    int log2[N],mi[N][25];
    void init(int n)
    {
        for(int i = 0; i <= n; i ++)log2[i] = (i == 0 ? -1 : log2[i >> 1] + 1);
        for(int j = 1; j < log2[n]; j ++)
            for(int i = 1; i + (1 << j) <= n + 1; i ++)
                mi[i][j] = std::min(mi[i][j - 1], mi[i + (1 << j - 1)][j - 1]);
    }
    int query(int ql, int qr)
    {
        int k = log2[qr - ql + 1];
        return std::min(mi[ql][k], mi[qr - (1 << k) + 1][k]);
    }
} rmq;
int sc(int i,int k,int len)
{
    if(k==1) return len-sa[i];
    return rmq.query(i+1,i+k-1);
}
int main()
{
    int t,k;std::cin>>t;
    while(t--)
    {
        int ans=0;
        scanf("%d%s",&k,ss);
        int len=strlen(ss);
        for(int i=0;i<len;i++)
            s[i]=ss[i]-'a'+1;
        s[len]=0;
        da(s,sa,len+1,28);
        calheight(s,sa,len);
        height[len+1]=0;
        for(int i=1;i<=len;i++)
            rmq.mi[i][0]=height[i];
        rmq.init(len);
        for(int i=1;i<=len-k+1;i++)
            ans+=std::max(sc(i,k,len)-std::max(height[i],height[i+k]),0);
        printf("%d\n",ans);
    }
    return 0;
}