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[];
 int ans;

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

     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 = ;
         len[] = ;
         memset(ch,,sizeof ch);
         memset(fa,,sizeof fa);
         memset(cnt,,sizeof cnt);
     }

     void add( int x)
     {
         int p = last, np = last = ++tot;
         len[np] = len[p] + , cnt[last] = ;
         while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
         if( p == )
             fa[np] = ;
         else
         {
             int q = ch[p][x];
             if( len[q] == len[p] + )
                 fa[np] = q;
             else
             {
                 int nq = ++tot;
                 memcpy( ch[nq], ch[q], sizeof ch[q]);
                 len[nq] = len[p] + , 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 = ; i <= len[last]; i++)   sum[i] = ;
         for(int i = ; i <= tot; i++)   sum[len[i]]++;
         for(int i = ; i <= len[last]; i++)   sum[i] += sum[i-];
         for(int i = ; 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=;
         sam.init();
         scanf("%d%s",&k,ss);
         for(int i=,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 ;
 }

后缀数组做法：

 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 #include <set>
 #include <iostream>
 const int N = ;
 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 = ; i < m; ++i) ws[i] = ;
     for (i = ; i < n; ++i) ws[x[i]=r[i]]++;
     for (i = ; i < m; ++i) ws[i] += ws[i-];
     for (i = n-; i >= ; --i) sa[--ws[x[i]]] = i;
     for (j = , p = ; p < n; j *= , m = p)
     {
         for (p = , i = n - j; i < n; ++i) y[p++] = i;
         for (i = ; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
         for (i = ; i < n; ++i) wv[i] = x[y[i]];
         for (i = ; i < m; ++i) ws[i] = ;
         for (i = ; i < n; ++i) ws[wv[i]]++;
         for (i = ; i < m; ++i) ws[i] += ws[i-];
         for (i = n-; i >= ; --i) sa[--ws[wv[i]]] = y[i];
         for (std::swap(x, y), p = , x[sa[]] = , i = ; i < n; ++i)
             x[sa[i]] = cmp(y, sa[i-], sa[i], j) ? p- : p++;
     }
 }

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