| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8277 | Accepted: 2369 |
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
題意：
給定一個字符串，從任意位置把它切爲兩半，得到兩條子串
定義 子串1爲s1，子串2爲s2，子串1的反串爲s3，子串2的反串爲s4
現在從s1 s2 s3 s4中任意取出兩個串組合，問有多少種不同的組合方法
分析：主要是哈希，一開始用的set，超時，最後用了一個不是太好的哈希
爛代碼
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <set>
using namespace std;
#define inf 40007
char str[500][80];
struct node
{
int g;
struct node *next;
};
struct node *h[50000],*p,*q,*r;
int top=0;
int ct=0;
void lianjie(char s1[],char s2[])
{
char ss1[100];
strcpy(ss1,s1);
int i,j;
char ss[100];
int ans=0;
strcpy(ss,strcat(ss1,s2));
int l=strlen(ss);
for(i=0;i<l;i++)
{
ans+=((ss[i]-'a'+1)*(i+1));
}
ans%=inf;
if(h[ans]==NULL)
{
strcpy(str[top],ss);
p=new node;
p->g=top;
p->next=NULL;
top++;
h[ans]=p;
ct++;
}
else
{
int flag=0;
q=h[ans];
while(q)
{
if(strcmp(str[q->g],ss)==0)
{
flag=1;
break;
}
q=q->next;
}
if(!flag)
{
q=h[ans];
while(q->next)
{
q=q->next;
}
strcpy(str[top],ss);
p=new node;
p->g=top;
p->next=NULL;
top++;
q->next=p;
ct++;
}
}
}
void f(int l,int r,char s[])
{
int i,j;
char s1[100],s2[100],s3[100],s4[100];
for(i=0;i<l;i++)
{
s1[i]=s[i];
s2[l-i-1]=s[i];
}
s1[l]=s2[l]='\0';
for(i=l;i<r;i++)
{
s3[i-l]=s[i];
s4[r-i-1]=s[i];
}
s3[r-l]=s4[r-l]='\0';
lianjie(s1,s3);
lianjie(s3,s1);
lianjie(s2,s3);
lianjie(s3,s2);
lianjie(s1,s4);
lianjie(s4,s1);
lianjie(s2,s4);
lianjie(s4,s2);
}
void init()
{
ct=0;
for(int i=0;i<=49999;i++)
{
h[i]=NULL;
}
top=0;
}
int main()
{
int n,m,i,j;
char s[100];
scanf("%d",&n);
while(n--)
{
init();
scanf("%s",s);
int l=strlen(s);
if(l==1)
{
printf("1\n");
}
else
{
for(i=1;i<l;i++)
{
f(i,l,s);
}
printf("%d\n",top);
}
}
return 0;
}
