ACM-BFS之Catch That Cow——hdu2717

Catch That Cow

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2717

Problem 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
4

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.

这道题本来很简单，叫我想复杂了，o(╯□╰)o囧啊。。。。

没想到暴力BFS就可以过。

我还在想 VIS数组是否有优化的选择，比如 用终点和当前点的距离差做标准等。。。

结果一直超内存。。。

后来，想干脆暴搜一次看看，

结果因为，我判断没写函数，直接判断，

if(vis[k.x]==0 && k.x>=0 && k.x<=100000)

因为if里的顺序问题，一直越界错误。。。（应该让判断范围在 判断数组内容前面）

然后又因为判断与0，100000 落了等号。。又WA一次。。

这个简单的题目，悲剧了好长时间。。。想哭了的说。。。

#include <iostream>
#include <queue>
#include <string.h>

using namespace std;
int finish;
bool vis[100011];
struct Coor
{
    int x,step;
};

bool judge(int x)
{
    if(x<0 || x>100000) return 0;
    if(vis[x]==1)   return 0;
    return 1;
}

int bfs(int start)
{
    memset(vis,0,sizeof(vis));
    queue <Coor> q;
    Coor t,k;

    vis[start]=1;
    t.x=start;
    t.step=0;

    q.push(t);

    while(!q.empty())
    {
        t=q.front();
        q.pop();

        if(t.x==finish) return t.step;

        k.x=t.x+1;
        if(judge(k.x))
        {
            k.step=t.step+1;
            vis[k.x]=1;
            q.push(k);
        }

        k.x=t.x-1;
        if(judge(k.x))
        {
            k.step=t.step+1;
            vis[k.x]=1;
            q.push(k);
        }

        k.x=t.x*2;
        if(judge(k.x))
        {
            k.step=t.step+1;
            vis[k.x]=1;
            q.push(k);
        }
    }
    return -1;
}

int main()
{
    int start;
    while(cin>>start>>finish)
    {
        cout<<bfs(start)<<endl;
    }
    return 0;
}