迭代加深搜索-POJ 3134 Power Calculus

文章目录

• 迭代加深搜索
• 例题
• • 题意
  • 分析
  • 代码
• 小结

迭代加深搜索

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什么是迭代加深搜索？ 迭代加深搜索(Iterative Deepening DFS,IDDFS)是一种结合了DFS和BFS思想的搜索方法。当搜索树很深且很宽的时候，用DFS会陷入递归无法返回，用BFS队列空间会爆炸，那么可以试试IDDFS，简单来说，就是每次限制搜索深度的DFS。比如DFS搜索k层，若没有找到可行解则立即返回，再DFS搜索k 1层，直到找到可行解为止，在层数上采用BFS思想来逐步扩大DFS的搜索深度。

IDA* 估价函数对迭代加深搜索的优化，即乐观估计剪枝。当找到解需要的至少层数 当前层数>层数限制时，直接退出。

例题

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传送门: POJ-3134

Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications: x2 = x × x, x3 = x2 × x, x4 = x3 × x, …, x31 = x30 × x. The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications: x2 = x × x, x3 = x2 × x, x6 = x3 × x3, x7 = x6 × x, x14 = x7 × x7, x15 = x14 × x, x30 = x15 × x15, x31 = x30 × x. This is not the shortest sequence of multiplications to compute x31. There are many ways with only seven multiplications. The following is one of them: x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x8 = x4 × x4, x10 = x8 × x2, x20 = x10 × x10, x30 = x20 × x10, x31 = x30 × x. If division is also available, we can find a even shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division): x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x16 = x8 × x8, x32 = x16 × x16, x31 = x32 ÷ x. This is one of the most efficient ways to compute x31 if a division is as fast as a multiplication. Your mission is to write a program to find the least number of operations to compute xn by multiplication and division starting with x for the given positive integer n. Products and quotients appearing in the sequence should be x to a positive integer’s power. In others words, x−3, for example, should never appear.

input:

The input is a sequence of one or more lines each containing a single integer n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.

output:

Your program should print the least total number of multiplications and divisions required to compute xn starting with x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.

Sample Input:

代码语言：javascript复制

1
31
70
91
473
512
811
953
0

Sample Output:

代码语言：javascript复制

0
6
8
9
11
9
13
12

题意

给定数x和n，求xn ，只能用乘法和除法，算过的结果可以被利用，问最少算多少次？

分析

等价于从数字1开始，用加减法，最少算多少次得到n? 用当前值和之前产生的值进行加减运算得到新的值，判断是否等于n。 估价函数：如果当前值用最快的方式（连乘2倍增）都不能到达n则退出。

代码

代码语言：javascript复制

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int maxn = 1010;
int a[maxn];
int n, k;
bool ida(int now,int curk) {
	if (curk > k)return true;//当前层数大于层数限制
	//剩下的层数(层数限制-当前层)用最乐观的倍增也不能达到n
	if (now << (k - curk) < n)return true;
	return false;
}
bool iddfs(int now, int curk) {
	if (ida(now, curk))return false;
	if (now == n)return true;
	a[curk] = now;
	for (int i = 0; i <= curk; i  ) {//遍历之前算过的值
		//加
		if (iddfs(now   a[i], curk   1))return true;
		//减
		else if (iddfs(abs(now - a[i]), curk   1))return true;
	}
	return false;
}
int main() {
	while (~scanf("%d", &n) && n) {
		for (k = 0;; k  ) {//每次最大搜索k层
			memset(a, 0, sizeof(a));
			if (iddfs(1, 0))break;//从数字1开始，当前层0
		}
		printf("%dn", k);
	}
	return 0;
}

小结

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1. 空间复杂度相对较小（其实前面k-1层的浪费对于复杂度来说可以忽略），如果广/深搜爆栈可以一试。
2. 适用于时间充裕，空间较小，没有明确搜索深度上限的题。
3. 剪枝!!!

原创不易，请勿转载（本不富裕的访问量雪上加霜 ） 博主首页：https://blog.csdn.net/qq_45034708

dfs division input integer sequence

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