初学算法----枚举总结

暴力枚举好像十分适用于---需要有限个且不太多的结果---的情况下:

这样我们就可以列有限的for循环逐一枚举全部情况;

如钟表问题:

有9个时钟，排成一个3*3的矩阵。

|-------|    |-------|    |-------|
|       |    |       |    |   |   |
|---O   |    |---O   |    |   O   |
|       |    |       |    |       |
|-------|    |-------|    |-------|
    A            B            C    

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O   |    |   O   |
|   |   |    |   |   |    |   |   |
|-------|    |-------|    |-------|
    D            E            F    

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O---|    |   O   |
|   |   |    |       |    |   |   |
|-------|    |-------|    |-------|
    G            H            I    
(图 1)

现在需要用最少的移动，将9个时钟的指针都拨到12点的位置。共允许有9种不同的移动。如下表所示，每个移动会将若干个时钟的指针沿顺时针方向拨动90度。

移动    影响的时钟
 
 1         ABDE
 2         ABC
 3         BCEF
 4         ADG
 5         BDEFH
 6         CFI
 7         DEGH
 8         GHI
 9         EFHI    

 

输入：

9个整数，表示各时钟指针的起始位置，相邻两个整数之间用单个空格隔开。其中，0=12点、1=3点、2=6点、3=9点。

输出：

输出一个最短的移动序列，使得9个时钟的指针都指向12点。按照移动的序号从小到大输出结果。相邻两个整数之间用单个空格隔开。

样例输入：

3 3 0 
2 2 2 
2 1 2 

样例输出：

4 5 8 9 
这里有有限个操作----9个,对钟表每次循环3次回到初始位置,钟表也只有9个;
则可以枚举9个操作的状态及次数,再判断对钟表的影响;
再如:

Safecracker

 1000ms  65536K

描述：

"The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. Fortunately old Brumbaugh from research knew Klein's secrets and wrote them down before he died. A Klein safe has two distinguishing features: a combination lock that uses letters instead of numbers, and an engraved quotation on the door. A Klein quotation always contains between five and twelve distinct uppercase letters, usually at the beginning of sentences, and mentions one or more numbers. Five of the uppercase letters form the combination that opens the safe. By combining the digits from all the numbers in the appropriate way you get a numeric target. (The details of constructing the target number are classified.) To find the combination you must select five letters v, w, x, y, and z that satisfy the following equation, where each letter is replaced by its ordinal position in the alphabet (A=1, B=2, ..., Z=26). The combination is then vwxyz. If there is more than one solution then the combination is the one that is lexicographically greatest, i.e., the one that would appear last in a dictionary."

v - w2+ x3- y4+ z5= target

"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 92+ 53- 34+ 25= 1. There are actually several solutions in this case, and the combination turns out to be LKEBA. Klein thought it was safe to encode the combination within the engraving, because it could take months of effort to try all the possibilities even if you knew the secret. But of course computers didn't exist then."


"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations.

输入：

Input consists of one or more lines containing a positive integer target less than twelve million, a space, then at least five and at most twelve distinct uppercase letters. The last line will contain a target of zero and the letters END; this signals the end of the input.

输出：

For each line output the unique Klein combination, or 'no solution' if there is no correct combination. Use the exact format shown below."

样例输入：

1 ABCDEFGHIJKL
11700519 ZAYEXIWOVU
3072997 SOUGHT
1234567 THEQUICKFROG
0 END

样例输出：

LKEBA
YOXUZ
GHOST
no solution
只要5个结果,5个for循环,枚举每一个可能的情况;