UVA 1374 迭代加深搜索
http://vjudge.net/contest/view.action?cid=50643#problem/C
Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:
x2 = x x x, x3 = x2 x x, x4 = x3 x x, ... , x31 = x30 x x.The operation of squaring can appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:
x2 = x x x, x3 = x2 x x, x6 = x3 x x3, x7 = x6 x x, x14 = x7 x x7,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:
x15 = x14 x x, x30 = x15 x x15, x31 = x30 x x.
x2 = x x x, x4 = x2 x x2, x8 = x4 x x4, x10 = x8 x x2,There however is no way to compute x31 with fewer multiplications. Thus this is one of the most efficient ways to compute x31 only by multiplications.
x20 = x10 x x10, x30 = x20 x x10, x31 = x30 x x.
If division is also available, we can find a shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division):
x2 = x x x, x4 = x2 x x2, x8 = x4 x x4, x16 = x8 x x8, x32 = x16 x 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 of operations should be x to a positive integer's power. In other 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
1 31 70 91 473 512 811 953 0
Sample Output
0 6 8 9 11 9 13 12
解题思路：最直接的想法就是利用dfs进行搜索，但是有一点我们在深搜的时候不知道什么时候是个结尾，因此我们可以利用限定搜索的层数。
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int limt_dept;
int n,num,vis[1020];
bool dfs()
{
if(num>limt_dept) return false;
if(vis[num]==n) return true;
if(vis[num]<<(limt_dept-num)<n)return false;
for(int i=0;i<=num;i++)
{
num++;
vis[num]=vis[num-1]+vis[i];
if(vis[num]<=2000&&dfs())return true;
vis[num]=vis[num-1]-vis[i];
if(vis[num]>0&&dfs())return true;
num--;
}
return false;
}
int main()
{
while(~scanf("%d",&n))
{
if(n==0)
break;
limt_dept=0;
while(limt_dept<=20)
{
num=0;
vis[num]=1;
if(dfs())
break;
limt_dept++;
}
printf("%d\n",limt_dept);
}
return 0;
}