POJ 3007 Organize Your Train part II(枚举)

Organize Your Train part II

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7531		Accepted: 2184

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

一个字符串，可以在它任意位置将它分成两部分，记为a1,a2

将a1 , a2 分别反过来 记为 a3 , a4 

任意将其中2个连接成一个新的字符串，统计组成的个数 （a1不能和a3连接,a2不能和a4）

可以的组合有  (a1 , a2) (a2 , a1)  (a1 , a4) (a4 , a1) 

            (a3 , a2) (a2 , a3)  (a3 , a4) (a4 , a3) 共八种

直接枚举出就可以了，用string + map写了好几遍，全部tle。。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <cmath>
#include <map>
#include <queue>
#define N 75
using namespace std;
struct node
{
   char a[N];
}q[N*N];
int cnt;
bool judge(char s[])
{
    if(cnt == 0) return true;
    for(int i=0;i<cnt;i++)
    {
        if(strcmp(q[i].a,s)==0)
            return false;
    }
    return true;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        char s[N];
        scanf("%s",s);
       int len=strlen(s);
       int ans=0;
       cnt=0;
       for(int i=0;i<len-1;i++)
       {

           char b1[80],b2[80];     // 如果四部分分别为 a1 a2 a3 a4
           int c1=0,c2=0;          // 其中 a3是a1的反串  a4是a2的反串
           for(int j=0;j<=i;j++)    //b1 = a1 + a4 
           {                        //b2 = a4 + a1
               b1[c1++]= s[j];    
           }
           for(int j=len-1;j>i;j--)
           {
               b1[c1++] = s[j];
               b2[c2++]= s[j];
           }
           for(int j=0;j<=i;j++)
           {
              b2[c2++] = s[j];
           }
           b1[c1]='\0';
           b2[c2]='\0';
           if(judge(b1))          //判断b1是否已经存在
           {
            strcpy(q[cnt++].a,b1);    //不存在记录下b1
               ans++;
           }
           if(judge(b2))
           {
             strcpy(q[cnt++].a,b2);
               ans++;
           }
           c1=0;
           c2=0;
           for(int j=0;j<=i;j++)    //b1 = a1 + a2
           {
              b1[c1++] = s[j];       //b2 = a2 + a1

           }
           for(int j=i+1;j<len;j++)
           {
               b1[c1++] = s[j];
               b2[c2++]= s[j];
           }
           for(int j=0;j<=i;j++)
           {
               b2[c2++]= s[j];
           }
          if(judge(b1))
           {
               strcpy(q[cnt++].a,b1);
               ans++;
           }
           if(judge(b2))
           {
             strcpy(q[cnt++].a,b2);
               ans++;
           }
           c1=0;
           c2=0;
           for(int j=i+1;j<len;j++)     //b1 = a2 + a3
           {
              b1[c1++] = s[j];          // b2 = a3 + a2
           }
           for(int j=i;j>=0;j--)
           {
               b1[c1++] = s[j];
               b2[c2++]= s[j];
           }
           for(int j=i+1;j<len;j++)
           {

               b2[c2++]= s[j];
           }
           if(judge(b1))
           {
               strcpy(q[cnt++].a,b1);
               ans++;
           }
           if(judge(b2))
           {
             strcpy(q[cnt++].a,b2);
               ans++;
           }
           c1=0;
           c2=0;
           for(int j=i;j>=0;j--)     //b1 = a3 + a4
           {
              b1[c1++] = s[j];       //b2 = a4 + a3

           }
           for(int j=len-1;j>i;j--)
           {
              b1[c1++] = s[j];
               b2[c2++]= s[j];
           }
           for(int j=i;j>=0;j--)
           {

               b2[c2++]= s[j];
           }
           if(judge(b1))
           {
               strcpy(q[cnt++].a,b1);
               ans++;
           }
           if(judge(b2))
           {
             strcpy(q[cnt++].a,b2);
               ans++;
           }
       }
       printf("%d\n",ans);
    }
    return 0;
}