Harry and magic string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 576 Accepted Submission(s): 287

Problem Description

Harry got a string T, he wanted to know the number of T’s disjoint palindrome substring pairs. A string is considered to be palindrome if and only if it reads the same backward or forward. For two substrings of

Input

There are several cases.
For each test case, there is a string T in the first line, which is composed by lowercase characters. The length of T is in the range of [1,100000].

Output

For each test case, output one number in a line, indecates the answer.

Sample Input

aca aaaa

Sample Output

3 15

Hint

For the first test case there are 4 palindrome substrings of T. They are: S1=T[0,0] S2=T[0,2] S3=T[1,1] S4=T[2,2] And there are 3 disjoint palindrome substring pairs. They are: (S1,S3) (S1,S4) (S3,S4). So the answer is 3.

Source

BestCoder Round #25

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题意：

在一个S串中找两个回文串，两个回文串没有重叠的部分。问有多少个这样的回文串对。

思路：

用manacher我们可以求出以$i$为中心的回文串的个数。

$pre[i]$表示以$i$为开头的回文串的个数，$suf[i]$表示以$i$为结尾的回文串的个数。

那么我们要求的答案其实就是$\sum_{i = 1}^{len} pre[i] * (\sum_{j=1}^{i-1} suf[j])$

所以我们再用一个数组$sum$来存$suf$数组的前缀和。

那么$pre$和$suf$怎么求呢？

我们manacher处理出的$p[i] /2$表示的是以$i$为中心的回文的个数。

那么对于$i$之前的半径内的字符，他们的$pre$可以$+1$。同样他之后的半径内的字符，$suf$可以$-1$

暴力统计肯定是不行的，我们可以使用差分的思想。

对于$[i,j]$区间内的每一个数都更新$k$的话，我们可以用一个数组$add[i] += k$,add[j+1] -=k$

然后从前往后求前缀和得到的就是对区间的全部更新了。求$suf$和$pre$也是同样的道理。

$pre$是后面影响前面的所以从后往前求“#”向前靠，$suf$从前往后求“#”向后靠。

 1 #include<iostream>
 2 //#include<bits/stdc++.h>
 3 #include<cstdio>
 4 #include<cmath>
 5 //#include<cstdlib>
 6 #include<cstring>
 7 #include<algorithm>
 8 //#include<queue>
 9 #include<vector>
10 //#include<set>
11 //#include<climits>
12 //#include<map>
13 using namespace std;
14 typedef long long LL;
15 #define N 100010
16 #define pi 3.1415926535
17 #define inf 0x3f3f3f3f
18 
19 const int maxn = 1e5 + 5;
20 char s[maxn], ss[maxn * 2];
21 LL pre[maxn * 2], suf[maxn * 2], sum[maxn * 2];
22 int lens, p[maxn * 2];
23 
24 int init()
25 {
26     ss[0] = '$';
27     ss[1] = '#';
28     int lenss = 2;
29     for(int i = 0; i < lens; i++){
30         ss[lenss++] = s[i];
31         ss[lenss++] = '#';
32     }
33     ss[lenss] = '\0';
34     return lenss;
35 }
36 
37 void manacher()
38 {
39     int lenss = init();
40     int id, mx = 0;
41     for(int i = 1; i < lenss; i++){
42         if(i < mx){
43             p[i] = min(p[2 * id - i], mx - i);
44         }
45         else{
46             p[i] = 1;
47         }
48         while(ss[i - p[i]] == ss[i + p[i]])p[i]++;
49         if(mx < i + p[i]){
50             id = i;
51             mx = i + p[i];
52         }
53 
54     }
55 }
56 
57 int main()
58 {
59     while(scanf("%s", s) != EOF){
60         lens = strlen(s);
61         /*for(int i = 0; i < lens * 2 + 3; i++){
62             p[i] = 0;
63         }*/
64         for(int i = 0; i < lens + 1; i++){
65             suf[i] = pre[i] = sum[i] = 0;
66         }
67 
68         manacher();
69 
70         for(int i = 2; i <= lens * 2; i++){
71             int x = (i + 1) / 2;//向上取整
72             suf[x]++, suf[x + (p[i] / 2)]--;
73         }
74         for(int i = lens * 2; i >= 2; i--){
75             int x = i / 2;
76             pre[x]++, pre[x - (p[i] / 2)]--;
77         }
78 
79         for(int i = lens; i >= 1; i--){
80             pre[i] += pre[i + 1];
81         }
82         for(int i = 1; i <= lens; i++){
83             suf[i] += suf[i - 1];
84             sum[i] += sum[i - 1] + suf[i];
85         }
86         LL ans = 0;
87         for(int i = 1; i <= lens; i++){
88             ans += (LL)pre[i] * sum[i - 1];
89         }
90         printf("%I64d\n", ans);
91     }
92     return 0;
93 }

hdu5157 Harry and magic string【manacher】

Harry and magic string

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