hdu5157 Harry and magic string【manacher】

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 T:x=T[a1…b1],y=T[a2…b2](where a1 is the beginning index of x,b1 is the ending index of x. a2,b2 as the same of y), if both x and y are palindromes and b1<a2 or b2<a1 then we consider (x, y) to be a disjoint palindrome substring pair of T.

 

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。

#include<iostream>
//#include<bits/stdc++.h>
#include<cstdio>
#include<cmath>
//#include<cstdlib>
#include<cstring>
#include<algorithm>
//#include<queue>
#include<vector>
//#include<set>
//#include<climits>
//#include<map>
using namespace std;
typedef long long LL;
#define N 100010
#define pi 3.1415926535
#define inf 0x3f3f3f3f

const int maxn = 1e5 + 5;
char s[maxn], ss[maxn * 2];
LL pre[maxn * 2], suf[maxn * 2], sum[maxn * 2];
int lens, p[maxn * 2];

int init()
{
    ss[0] = '$';
    ss[1] = '#';
    int lenss = 2;
    for(int i = 0; i < lens; i++){
        ss[lenss++] = s[i];
        ss[lenss++] = '#';
    }
    ss[lenss] = '';
    return lenss;
}

void manacher()
{
    int lenss = init();
    int id, mx = 0;
    for(int i = 1; i < lenss; i++){
        if(i < mx){
            p[i] = min(p[2 * id - i], mx - i);
        }
        else{
            p[i] = 1;
        }
        while(ss[i - p[i]] == ss[i + p[i]])p[i]++;
        if(mx < i + p[i]){
            id = i;
            mx = i + p[i];
        }

    }
}

int main()
{
    while(scanf("%s", s) != EOF){
        lens = strlen(s);
        /*for(int i = 0; i < lens * 2 + 3; i++){
            p[i] = 0;
        }*/
        for(int i = 0; i < lens + 1; i++){
            suf[i] = pre[i] = sum[i] = 0;
        }

        manacher();

        for(int i = 2; i <= lens * 2; i++){
            int x = (i + 1) / 2;//向上取整
            suf[x]++, suf[x + (p[i] / 2)]--;
        }
        for(int i = lens * 2; i >= 2; i--){
            int x = i / 2;
            pre[x]++, pre[x - (p[i] / 2)]--;
        }

        for(int i = lens; i >= 1; i--){
            pre[i] += pre[i + 1];
        }
        for(int i = 1; i <= lens; i++){
            suf[i] += suf[i - 1];
            sum[i] += sum[i - 1] + suf[i];
        }
        LL ans = 0;
        for(int i = 1; i <= lens; i++){
            ans += (LL)pre[i] * sum[i - 1];
        }
        printf("%I64d
", ans);
    }
    return 0;
}