UVa11567 - Moliu Number Generator(记忆化搜索dfs)

最新推荐文章于 2018-04-04 07:57:40 发布

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本文介绍了一个通过递归和记忆化搜索解决的问题：如何用最少的操作步骤将初始值0转换为目标整数S。允许的操作包括加1、减1和翻倍。文章提供了完整的C++实现代码，展示了如何有效地计算最小操作数。

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ProblemC: Moliu Number Generator

Let's play a number game. We start with N = 0, and we want to make N = a given integer S.

Only three types of operations are allowed:

INC : increment N by 1, i.e. N <-- N + 1
DEC : decrement N by 1, i.e. N <-- N - 1
DBL : double N, i.e. N <-- 2 N

Of course we want to make N = S with the minimum number of operations. Consider an example: Let S = 7. Then only 5 steps are required, for instance:

INC : N = 0 + 1 = 1
INC : N = 1 + 1 = 2
DBL : N = 2 × 2 = 4
DBL : N = 2 × 4 = 8
DEC : N = 8 - 1 = 7 <-- DONE!!

Input

Input contains no more than 200 lines. Each line contains one integer S (0 ≤ S ≤ 2³¹). Input is terminated by EOF.

Output

For each S, output the minimum number of operations required to make N = S. You may assume that N is of infinite precision, so NO overflow will ever occur.

Sample Input

Sample Output

记忆化搜索

#include <cstdio>
#include <algorithm>
#include <map>

using namespace std;

long long n;
map<long long, int> m;

bool input();
void solve();
int f(long long n);

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("d:\\OJ\\uva_in.txt", "r", stdin);
	#endif

	while (input()) {
		solve();
	}
	return 0;
}

bool input()
{
	return scanf("%lld", &n) == 1;
}

int f(long long n)
{
	if (m.count(n)) return m[n];
	
	if (n == 0) return 0;
	
	if (n == 1) return 1;
	
	if (!(n & 1)) return m[n] = f(n / 2) + 1;
	else {
		return m[n] = min(f(n + 1), f(n - 1)) + 1;
	}
}

void solve()
{
	printf("%d\n", f(n));
}