poj 3134 分类: poj 2015-08-08 23:42 4人阅读 评论(0) 收藏
喜闻乐见的迭代加深搜索
快速幂最多只需要
然而这道题的答案会更小,迭代加深搜索显然是可以承受的。
剪枝优化:
如果剩余的操作次数都用来将最大数与最大数相乘,
得到的结果仍然比希望得到的数小,那么这个状态就一定是不可行的
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <iostream>
#include <algorithm>
template<class Num>void read(Num &x)
{
char c; int flag = 1;
while((c = getchar()) < '0' || c > '9')
if(c == '-') flag *= -1;
x = c - '0';
while((c = getchar()) >= '0' && c <= '9')
x = (x<<3) + (x<<1) + (c-'0');
x *= flag;
return;
}
template<class Num>void write(Num x)
{
if(x < 0) putchar('-'), x = -x;
static char s[20];int sl = 0;
while(x) s[sl++] = x%10 + '0',x /= 10;
if(!sl) {putchar('0');return;}
while(sl) putchar(s[--sl]);
}
const int size = 2000;
int target, depth;
int mass[size];
bool dfs(int pos)
{
if(pos > depth || ((mass[pos] << (depth - pos))) < target) return false;
if(mass[pos] == target) return true;
for(int i = 1; i <= pos; i++)
{
if(mass[pos] + mass[i] <= size)
{
mass[pos + 1] = mass[pos] + mass[i];
if(dfs(pos + 1)) return true;
}
if(mass[pos] - mass[i] > 0)
{
mass[pos + 1] = mass[pos] - mass[i];
if(dfs(pos + 1)) return true;
}
}
return false;
}
void solve()
{
depth = 1, mass[1] = 1;
while(true)
{
if(dfs(1))
{
depth--;
return;
}
depth++;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("3134.in","r",stdin);
freopen("3134.out","w",stdout);
#endif
while(true)
{
read(target);
if(!target) break;
solve();
write(depth), puts("");
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
#endif
return 0;
}
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