hdu 5157(树状数组+Manacher)

Harry and magic string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 236 Accepted Submission(s): 118


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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].

[align=left]Output[/align]
For each test case, output one number in a line, indecates the answer.

[align=left]Sample Input[/align]

aca
aaaa

[align=left]Sample Output[/align]

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.

[align=left]Source[/align]
BestCoder Round #25

题意:找到一个串中不相交的回文串的数量.
题解:官方题解:
先求出p[i],表示以i为回文中心的回文半径，manacher，sa，hash都可以。然后我们考虑以i为回文中心的时候，可以贡献出多少答案。我们 先考虑长度为p[i]的那个回文子串，它能贡献的答案，就是末尾在i-p[i]之前的回文子串数，那么长度为p[i-1]的，也是同样的道理。所以我们需 要求一个f[x]，表示末尾不超过x的回文子串总共有多少个，把f[i-p[i]]到f[i-1]累加起来即为以i为中心的回文子串能贡献的答案。那我们 先统计，以x为结尾的回文子串有多少个，设为g[x]。来看以j为中心的回文子串，它的回文半径是p[j]，那么从j到j+p[j]-1的这一段，都会被 贡献一个回文子串，也就是这一段的g[]会被贡献1，这里的处理方法很多，不细说。然后求一次前缀和，即可得到f[x]。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
#include <queue>
using namespace std;
typedef long long LL;
const int N = 100005;
char s[2*N],t[2*N];
char str[2*N];
int p[2*N];
int c[2*N];
LL l[2*N];
void manacher(int n)
{
int id =0,maxr=0;
p[0]=1;
for(int i=1; i<2*n+1; i++)
{
if(p[id]+id>i)
p[i]=min(p[2*id-i],p[id]+id-i);
else p[i]=1;
while(s[i-p[i]]==s[i+p[i]])
++p[i];
if(id+p[id]<i+p[i]) id=i;
if(maxr<p[i])
maxr=p[i];
}
}
int lowbit(int x)
{
return x&-x;
}
void update(int idx,int v)
{
for(int i=idx; i<=2*N; i+=lowbit(i))
{
c[i]+=v;
}
}
int getsum(int idx)
{
int sum = 0;
for(int i=idx; i>0; i-=lowbit(i))
{
sum+=c[i];
}
return sum;
}
int main()
{
while(scanf("%s",str)!=EOF)
{
int n=strlen(str);
s[0]='#';
for(int i=1; i<=n; i++)
{
s[2*i]='#';
s[2*i-1]=str[i-1];
}
manacher(2*n);
memset(c,0,sizeof(c));
for(int i=1; i<=2*n; i++)
{
int ori;
if(i%2==0) ori = i/2+1;
else ori = (i+1)/2;
int ori1 = (i+p[i]-1)/2+1;
if(ori1<=ori) continue;
update(ori,1);
update(ori1,-1);
}
memset(l,0,sizeof(l));
for(int i=1; i<=n; i++)
{
l[i] = l[i-1]+getsum(i);
}
reverse(s,s+2*n+1);
manacher(2*n);
memset(c,0,sizeof(c));
for(int i=1; i<=2*n; i++)
{
int ori;
if(i%2==0) ori = i/2+1;
else ori = (i+1)/2;
int ori1 = (i+p[i]-1)/2+1;
if(ori1<=ori) continue;
update(ori,1);
update(ori1,-1);
}
LL ans = 0;
for(int i=1; i<=n; i++)
{
ans += getsum(i)*l[n-i];
}
printf("%lld\n",ans);
}
return 0;
}