poj 3278 Catch That Cow

[align=center]Catch That Cow[/align]

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 68738		Accepted: 21645

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.
Source
USACO 2007 Open Silver

一道简单的bfs题，只是注意要剪枝；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#include<set>
#include<functional>
#define pi 3.14159265358979323846
using namespace std;
int N,K;
struct point
{
int x;
int step;
};
bool vis[100005];
int bfs()
{
memset(vis,0,sizeof(vis));
point temp;
temp.x=N;
temp.step=0;
vis
=1;
queue<point> q;
q.push(temp);
point a,b,c;
while(!q.empty())
{
temp=q.front();
q.pop();
if(temp.x==K) return temp.step;
a.x=temp.x-1;
a.step=temp.step+1;
if(a.x>=0&&a.x<=100000&&vis[a.x]==0)
{
vis[a.x]=1;
q.push(a);

}
b.x=temp.x+1;
b.step=temp.step+1;
if(b.x>=0&&b.x<=100000&&vis[b.x]==0)
{
vis[b.x]=1;
q.push(b);
}
c.x=temp.x*2;
c.step=temp.step+1;
if(c.x>=0&&c.x<=100000&&vis[c.x]==0)
{
vis[c.x]=1;
q.push(c);
}
}
}
int main()
{
scanf("%d %d",&N,&K);
if(N>=K)
{
printf("%d\n",N-K);
}
else
{
int ans=bfs();
printf("%d\n",ans);
}
return 0;
}