POJ 3278 抓牛简单广搜

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 75331		Accepted: 23781

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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.

如Hint中所说，在一条线上抓，输入的N是人的起始位置，K是牛的人要到达的位置

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 100050;

int main()
{
    queue<int> a;
    int vis[maxn];
    int n,k;
    cin>>n>>k;

    memset(vis,0,sizeof(vis));
    int flag = 0;
    vis[n]=1;
    int step = 0;

    a.push(n);

if(n!=k){  //要考虑这种情况，不然0步输出的是1步
while(!a.empty()&&!flag){
            int t = a.size();//队列中元素的个数
            step++;
    while(t--){ //参考比较模板的while(head<tail),这里直接用的STL,所以形式变了
            int next,nnext[3];
            next = a.front();
            nnext[0]=next+1;
            nnext[1]=next-1;
            nnext[2]=next*2;

            a.pop();
            for(int i = 0;i<3;i++){
                if(nnext[i]==k){
                        flag=1;
                        break;}

                    if(nnext[i]>=1&&nnext[i]<=100000&&!vis[nnext[i]]){
                        vis[nnext[i]]=1;
                        a.push(nnext[i]);
                    }
                }
                if(flag) break;//我觉得这一步能省时间的，可是去掉不去掉提交的时间都一样，可能数据比较小吧 
                               //当n=1,k=2时，在step=1时，进栈的顺序是2,0，0，找到第一个2直接退出，因为此时step都是1
            }
    }

}


cout<<step<<endl;
return 0;
}