Hesty Str1ng

Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)

A chrysanthemum was painted on the second page, and we tried to use the magic power learned just now.

The beautiful flower was compacted to a colorful string

As known, if we try to choose a substring

The poetry told us the key to the eternity living secret is the number of the distinct palindrome strings made by the method above.

Two strings are considered different if and only if there are different characters in the same position of two strings.

Input

Only one line contains the string

Output

The key to the eternity living secret.

Sample input and output

Sample Input	Sample Output
abc	3

Hint

The palindrome strings can be made are "aba", "bcb", "abcba".

思路：

　　对于一个长度为n的子串，如果在其后连接一个长度为x（1<=x<n）的字符串形成新串T，并且T为回文串。

　　n=1时：形成的T的数量=0

　　n>1时：形成的T的数量=1+sum.（sum：子串含有的不同回文后缀的数量）

　　回顾下计算不同子串个数的后缀数组做法：

　　下面先给出一个结论：

　　　　sum[x]:表示后缀s[x....n-1]中s[k]==s[k+1]的个数

　　　　ans=∑（n-sa[i]-height[i]+sum[sa[i]+height[i]]) （字符串下标从0开始。）

　　　　并且当

　　　　　　　height[i]==0时　　ans-=1;

　　　　　　　height[i]>=2&&ss[sa[i]+height[i]-1]==ss[sa[i]+height[i]]时　　ans+=1;

　　　　　　　!height[i]&&ss[sa[i]+height[i]]==ss[sa[i]+height[i]+1]时　　ans-=1;

　　　　上面对应的三种情况分别是：

　　　　1. 此时有排序后的后缀abbb,ba.

　　　　2. 此时有排序后的后缀abb,abbba

　　　　3. 此时有排序后的后缀a,bba

　　具体证明过程我就不写了（PS：其实是我也不太会）

　　参考自校队另一位dalao的博文：http://blog.csdn.net/prolightsfxjh/article/details/66970491

　　具体见代码

 1 #include <cstdlib>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 const int N = 100005;
 7 int wa[N], wb[N], ws[N], wv[N];
 8 int s[N],sa[N],rank[N], height[N];
 9 char ss[N];
10 int sum[N];
11 bool cmp(int r[], int a, int b, int l)
12 {
13     return r[a] == r[b] && r[a+l] == r[b+l];
14 }
15 
16 void da(int r[], int sa[], int n, int m)
17 {
18     int i, j, p, *x = wa, *y = wb;
19     for (i = 0; i < m; ++i) ws[i] = 0;
20     for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
21     for (i = 1; i < m; ++i) ws[i] += ws[i-1];
22     for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
23     for (j = 1, p = 1; p < n; j *= 2, m = p)
24     {
25         for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
26         for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
27         for (i = 0; i < n; ++i) wv[i] = x[y[i]];
28         for (i = 0; i < m; ++i) ws[i] = 0;
29         for (i = 0; i < n; ++i) ws[wv[i]]++;
30         for (i = 1; i < m; ++i) ws[i] += ws[i-1];
31         for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
32         for (std::swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
33             x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
34     }
35 }
36 
37 void calheight(int r[], int sa[], int n)
38 {
39     int i, j, k = 0;
40     for (i = 1; i <= n; ++i) rank[sa[i]] = i;
41     for (i = 0; i < n; height[rank[i++]] = k)
42         for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
43 }
44 
45 int main()
46 {
47     int len;
48     long long ans=0;
49     scanf("%s",ss);
50     len=strlen(ss);
51     for(int i=0;i<len;i++)
52         s[i]=ss[i]-'a'+1;
53     s[len]=0;
54     da(s,sa,len+1,28);
55     calheight(s,sa,len);
56     for(int i=len-1;i>=0;i--)
57     if(ss[i]==ss[i+1])  sum[i]=sum[i+1]+1;
58     else    sum[i]=sum[i+1];
59     for(int i=1;i<=len;i++)
60     {
61         if(height[i]==0)ans--;
62         ans+=sum[sa[i]+height[i]]+len-sa[i]-height[i];
63         if(height[i]>=2&&ss[sa[i]+height[i]-1]==ss[sa[i]+height[i]])
64             ans++;
65         if(!height[i]&&ss[sa[i]+height[i]]==ss[sa[i]+height[i]+1])
66             ans--;
67         //printf("x==%d %lld\n",i,ans);
68     }
69     printf("%lld\n",ans);
70     return 0;
71 }