poj3278 Catch That Cow(简单的一维bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 47010		Accepted: 14766

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.

数组必须开大点

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

using namespace std;
struct node
{
    int x,ans;
} q[1000005];
int jx[]= {-1,1};
int n,k;
int map[1000005],v[1000005];
void bfs()
{
    struct node t,f;
    int e=0,s=0;
    t.x=n;
    v[t.x]=1;
    t.ans=0;
    q[e++]=t;
    while(s<e)
    {
        t=q[s++];
        if(t.x==k)
        {
            printf("%d
",t.ans);
            break;
        }
        for(int i=0; i<3; i++)
        {
            if(i==2)
                f.x=t.x*2;
            else f.x=t.x+jx[i];
            if(!v[f.x]&&f.x>=0&&f.x<=100000)
            {
                f.ans=t.ans+1;
                q[e++]=f;
                v[f.x]=1;
            }
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(map,0,sizeof(map));
        memset(v,0,sizeof(v));
        bfs();
    }
    return 0;
}