♥POJ 3278-Catch That Cow【搜索】-蒲公英云

♥POJ 3278-Catch That Cow【搜索】

C - Catch That Cow

Crawling in process… Crawling failed

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status Practice POJ 3278

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

解题思路：

题目大意就是有一个农夫和一头牛，在n，k，农夫想要到牛哪里，有两种移动方式，第一是坐标加1，第二是坐标乘2。注意搜索时超内存。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm> 
using namespace std;
struct node
{
    int x,step;
};
int flag[100005];
bool judge(node p)
{
    if(p.x<0||p.x>100000||p.step>flag[p.x])//判断到这一点时间是不是最短的 是否更新 
    return false;
    return true;
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==k)
        {
            printf("0\n");
            continue;
        }
        memset(flag,0x7f,sizeof(flag));
        node ks;
        ks.x=n;
        ks.step=0;
        flag[ks.x]=0;
        queue<node>q;
        q.push(ks);
        int wc=0;
        while(!q.empty())
        {
            node end,end2;
            end2=q.front();
            q.pop();
            for(int i=0;i<3;i++)
            {
                end=end2;
                if(i==0)
                {
                    end.x++;
                }
                if(i==1)
                {
                    end.x*=2;
                }
                if(i==2)
                {
                    end.x--;
                }
                end.step=end2.step+1;
                if(!judge(end))//如果不判断一下   直接进队是会超内存  各种超内存 
                {
                    continue;
                }
                if(end.x==k)
                {
                    printf("%d\n",end.step);
                    wc=1;
                    break;
                }
                flag[end.x]=end.step;
                q.push(end);
            }
            if(wc==1)
            break;
        }
    } 
    return 0;
}