poj 3278 Catch That Cow(bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 54700		Accepted: 17101

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

再次应证了我刚开始接触搜索时看到的一句话：搜索很强大的！

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<algorithm>

#include<queue>

using namespace std;

int N,K,vis[1000010];

const int M =1000000;

struct node{

int x,step;

};

int check(int x)

{

if(x<0 || x>=M||vis[x])return 0; //这里好像要把判边界放在前面，否则会爆。。可能判边界的情况更多

return 1;

}

int bfs(int x)

{

queue<node>Q;

int i;

node p,a;

p.x=x;

p.step=0;

vis[x]=1;

Q.push(p);

while(!Q.empty())

{

p=Q.front();

Q.pop();

if(p.x==K)return p.step;

a=p;

a.x=p.x+1;

if(check(a.x))

{

a.step=p.step+1;

vis[a.x]=1;

Q.push(a);

}

a.x=p.x-1;

if(check(a.x)){

a.step=p.step+1;

vis[a.x]=1;

Q.push(a);

}

a.x = p.x*2;

if(check(a.x)){

a.step=p.step+1;

vis[a.x]=1;

Q.push(a);

}

}

return 0;

}

int main()

{

int ans;

while(scanf("%d%d",&N,&K)!=EOF)

{

memset(vis,0,sizeof(vis));

ans=bfs(N);

printf("%d\n",ans);

}

return 0;

}