string string string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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.
∑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
Solution
题意：给一个串，问你其中正好出现k次的子串有多少个
思路：求出给的串的后缀数组和height数组，然后对于height数组中一个长度k-1的区间，如果两边的值比区间内的最小值（可以RMQ进行O(1)的查询）都要小，那这段区间就是对答案有贡献的。注意k=1时需要特殊处理一下，此时就是后缀数组派上用场的时候了。有不少细节在里面，具体请看代码。还有一个坑到哭泣的地方：由于是多组数据，一定要记得每次清空height数组（不清空的话如果这次的字符串比上一次的短，访问height[n+1]时预期的是0，但实际上是上次字符串的height值），否则要找十年才知道WA在哪。
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define MAXN 101111
4 using namespace std;
5 char s[MAXN];
6 int t1[MAXN],t2[MAXN],cc[MAXN],sa[MAXN],rnk[MAXN],height[MAXN];
7 int len;
8 bool cmp(int *y,int a,int b,int k)
9 {
10 int a1=y[a];
11 int b1=y[b];
12 int a2=a+k>=len ? -1:y[a+k];
13 int b2=b+k>=len ? -1:y[b+k];
14 return a1==b1 && a2==b2;
15 }
16 int make_sa()
17 {
18 int *x=t1,*y=t2;
19 int m=30;
20 for(int i=0; i<m; i++) cc[i]=0;
21 for(int i=0; i<len; i++) ++cc[x[i]=s[i]-‘a‘];
22 for(int i=1; i<m; i++) cc[i]+=cc[i-1];
23 for(int i=len-1; i>=0; i--) sa[--cc[x[i]]]=i;
24
25 for(int k=1; k<=len; k<<=1)
26 {
27 int p=0;
28 for(int i=len-k; i<len; i++) y[p++]=i;
29 for(int i=0; i<len; i++)
30 if( sa[i]>=k ) y[p++]=sa[i]-k;
31
32 for(int i=0; i<m; i++) cc[i]=0;
33 for(int i=0; i<len; i++) ++cc[x[y[i]]];
34 for(int i=1; i<m; i++) cc[i]+=cc[i-1];
35 for(int i=len-1; i>=0; i--) sa[--cc[x[y[i]]]]=y[i];
36
37 swap(x,y);
38 m=1; x[sa[0]]=0;
39 for(int i=1; i<len; i++)
40 x[sa[i]]=cmp(y,sa[i],sa[i-1],k) ? m-1:m++;
41
42 if( m>=len ) break;
43 }
44 }
45 void make_height()
46 {
47 for(int i=0; i<len; i++) rnk[sa[i]]=i;
48 height[0]=0;
49 int k=0;
50 for(int i=0; i<len; i++)
51 {
52 if(!rnk[i]) continue;
53 int j=sa[rnk[i]-1];
54 if(k) k--;
55 while(s[i+k]==s[j+k]) k++;
56 height[rnk[i]]=k;
57 }
58 }
59
60
61 int n,minl[100022][18];
62
63 void S_table(){
64 int l = log(len) / log(2);
65 for(int j = 1;j <= l;++j){
66 for(int i = 1;i + (1 << (j - 1)) - 1 <= len;++i){
67 minl[i][j] = min(minl[i][j-1],minl[i+(1<<(j-1))][j-1]);
68 }
69 }
70 }
71
72 int rmq(int l,int r){
73 int k = log(r-l+1) / log(2);
74 return min(minl[l][k],minl[r-(1<<k)+1][k]);
75 }
76
77
78 int T,k,cnt[30];
79
80 int main(){
81 cin >> T;
82 while(T--){
83 int ans = 0;
84 memset(height,0,sizeof height); /**< !!! */
85 scanf("%d",&k);
86 scanf("%s",s); len = strlen(s);
87 make_sa();
88 make_height();
89 n = len - 1;
90 for(int i = 0;i < 100022;++i) minl[i][0] = 1e9;
91 for(int i = 1;i <= len;++i)
92 minl[i][0] = height[i];
93 S_table();
94
95
96 if(k == 1){
97 for(int i = 1;i <= n+1;++i){
98 int tmp = len - sa[i-1];
99 if(height[i-1] < tmp && height[i] < tmp)
100 ans += tmp - max(height[i-1],height[i]);
101 }
102 printf("%d\n",ans);
103 continue;
104 }
105
106 for(int i = 1;i <= n - k + 2;++i){
107 int tmp = rmq(i,i+k-2);
108 if(height[i-1] < tmp && height[i+k-1] < tmp)
109 ans += tmp - max(height[i-1],height[i+k-1]);
110 }
111
112 printf("%d\n",ans);
113 }
114 }