POJ-3007 Organize Your Train part II （字符串hash）

Description

RJ Freight, a Japanese railroad company for freight operations has recently constructed exchange lines at Hazawa, Yokohama. The layout of the lines is shown in Figure 1.



Figure 1: Layout of the exchange lines
A freight train consists of 2 to 72 freight cars. There are 26 types of freight cars, which are denoted by 26 lowercase letters from "a" to "z". The cars of the same type are indistinguishable from each other, and each car's direction doesn't matter either.
Thus, a string of lowercase letters of length 2 to 72 is sufficient to completely express the configuration of a train.

Upon arrival at the exchange lines, a train is divided into two sub-trains at an arbitrary position (prior to entering the storage lines). Each of the sub-trains may have its direction reversed (using the reversal line). Finally, the two sub-trains are connected
in either order to form the final configuration. Note that the reversal operation is optional for each of the sub-trains.

For example, if the arrival configuration is "abcd", the train is split into two sub-trains of either 3:1, 2:2 or 1:3 cars. For each of the splitting, possible final configurations are as follows ("+" indicates final concatenation position):

[3:1]
abc+d  cba+d  d+abc  d+cba
[2:2]
ab+cd  ab+dc  ba+cd  ba+dc  cd+ab  cd+ba  dc+ab  dc+ba
[1:3]
a+bcd  a+dcb  bcd+a  dcb+a

Excluding duplicates, 12 distinct configurations are possible.

Given an arrival configuration, answer the number of distinct configurations which can be constructed using the exchange lines described above.

Input

The entire input looks like the following.

the number of datasets = m

1st dataset

2nd dataset

...
m-th dataset

Each dataset represents an arriving train, and is a string of 2 to 72 lowercase letters in an input line.

Output

For each dataset, output the number of possible train configurations in a line. No other characters should appear in the output.

Sample Input

4
aa
abba
abcd
abcde

Sample Output

1
6
12
18

Source
Japan 2006 Domestic

题意：将一个字符串可以任意分成两节，任意颠倒顺序，问最终会得到几个字符串。

分析：字符串hash，我没有进行查找判重操作，直接开了一个unsigned ll的set去记录出现过的值。

#include <set>
#include <queue>
#include <vector>
#include <cstdio>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
char S[100005];
unsigned long long xl[100005],Hash[100005],Rhash[100005];
int T;
int main()
{
xl[0] = 1;
for(int i = 1;i <= 100000;i++) xl[i] = xl[i-1]*27;
scanf("%d",&T);
while(T--)
{
int ans = 1;
set <unsigned long long> f;
scanf("%s",S);
int n = strlen(S);
Hash
= 0;
Rhash[0] = S[0];
for(int i = n-1;i >= 0;i--) Hash[i] = S[i] + Hash[i+1]*27;
for(int i = 1;i < n;i++) Rhash[i] = S[i] + Rhash[i-1]*27;
f.insert(Hash[0]);
for(int i = 0;i < n-1;i++)
{
if(!f.count(Rhash[i]+Hash[i+1]*xl[i+1]))
{
f.insert(Rhash[i]+Hash[i+1]*xl[i+1]);
ans++;
}
if(!f.count(Rhash[i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])*xl[i+1]))
{
f.insert(Rhash[i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])*xl[i+1]);
ans++;
}
if(!f.count(Hash[0]-Hash[i+1]*xl[i+1]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])*xl[i+1]))
{
f.insert(Hash[0]-Hash[i+1]*xl[i+1]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])*xl[i+1]);
ans++;
}
if(!f.count(Rhash[i]*xl[n-1-i]+Hash[i+1]))
{
f.insert(Rhash[i]*xl[n-1-i]+Hash[i+1]);
ans++;
}
if(!f.count(Rhash[i]*xl[n-1-i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])))
{
f.insert(Rhash[i]*xl[n-1-i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i]));
ans++;
}
if(!f.count((Hash[0]-Hash[i+1]*xl[i+1])*xl[n-1-i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i])))
{
f.insert((Hash[0]-Hash[i+1]*xl[i+1])*xl[n-1-i]+(Rhash[n-1]-Rhash[i]*xl[n-1-i]));
ans++;
}
if(!f.count((Hash[0]-Hash[i+1]*xl[i+1])*xl[n-1-i]+Hash[i+1]))
{
f.insert((Hash[0]-Hash[i+1]*xl[i+1])*xl[n-1-i]+Hash[i+1]);
ans++;
}
}
if(n > 1) cout<<ans<<endl;
else cout<<0<<endl;
}
}