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RE了n次，终于过了，贴代码，大家注意使用队列时范围的问题~~~

Posted by yingxiang720 at 2011-03-22 14:05:57 on Problem 3278

#include <iostream>
#include <queue>
using namespace std;

int n,k;
int number[210000];
int visit[210000];

int main()
{
    while(scanf("%d%d",&n,&k) != EOF)
    {
        queue<int>q;
        memset(number,0,sizeof(number));
        memset(visit,0,sizeof(visit));
        q.push(n);
        number[n] ++;
        visit[n] ++;
        while(!q.empty())
        {
            int cf = q.front();
            if(cf == k) break;
            q.pop();
            if(!visit[cf - 1] && cf - 1 <= 100000 && cf - 1 >= 0)
            {
                q.push(cf - 1);
                visit[cf - 1] ++;
                number[cf - 1] += number[cf] + 1;
            }
            if(!visit[cf + 1] && cf + 1 <= 100000 && cf + 1 >= 0)
            {
                q.push(cf + 1);
                visit[cf + 1] ++;
                number[cf + 1] += number[cf] + 1;
            }
            if(!visit[2 * cf] && cf * 2 <= 100000 && cf * 2 >= 0)
            {
                q.push(2 * cf);
                visit[2 * cf] ++;
                number[2 * cf] += number[cf] + 1;
            }
        }
        while(!q.empty())   q.pop();
        printf("%d\n",number[k] - 1);
    }
    return 0;
}

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要检查越界～～ (0) BUPT_ERIC 2011-05-05 18:17:50

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Content:

> #include <iostream>
> #include <queue>
> using namespace std;
> 
> int n,k;
> int number[210000];
> int visit[210000];
> 
> int main()
> {
>     while(scanf("%d%d",&n,&k) != EOF)
>     {
>         queue<int>q;
>         memset(number,0,sizeof(number));
>         memset(visit,0,sizeof(visit));
>         q.push(n);
>         number[n] ++;
>         visit[n] ++;
>         while(!q.empty())
>         {
>             int cf = q.front();
>             if(cf == k) break;
>             q.pop();
>             if(!visit[cf - 1] && cf - 1 <= 100000 && cf - 1 >= 0)
>             {
>                 q.push(cf - 1);
>                 visit[cf - 1] ++;
>                 number[cf - 1] += number[cf] + 1;
>             }
>             if(!visit[cf + 1] && cf + 1 <= 100000 && cf + 1 >= 0)
>             {
>                 q.push(cf + 1);
>                 visit[cf + 1] ++;
>                 number[cf + 1] += number[cf] + 1;
>             }
>             if(!visit[2 * cf] && cf * 2 <= 100000 && cf * 2 >= 0)
>             {
>                 q.push(2 * cf);
>                 visit[2 * cf] ++;
>                 number[2 * cf] += number[cf] + 1;
>             }
>         }
>         while(!q.empty())   q.pop();
>         printf("%d\n",number[k] - 1);
>     }
>     return 0;
> }

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