poj3278

Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 22900 Accepted: 7027 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

 

简单BFS，注意检查达到的位置是否越界了，re了两次。。。

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

using namespace std;

const int SIZE=210000;

typedef struct
{
    int x;
    int time;
}Loc;

queue<Loc> lqueue;
bool visit[SIZE];

int main()
{
    Loc temp, temp1;
    int n,k;
    int min_time;

    while(cin>>n>>k)
    {
        memset(visit,0,sizeof(visit));
        temp.x=n;
        temp.time=0;
        visit[temp.x]=true;
        lqueue.push(temp);
        while(!lqueue.empty())
        {
            temp=lqueue.front();
            lqueue.pop();

            if(temp.x==k)//find
            {
                min_time=temp.time;
                break;
            }

            temp1.time=temp.time+1;
            //W op
            temp1.x=temp.x+1;
            if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)  //检查是否越界
            {
                visit[temp1.x]=true;
                lqueue.push(temp1);
            }

            temp1.x=temp.x-1;
            if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)
            {
                visit[temp1.x]=true;
                lqueue.push(temp1);
            }
            //T op
            temp1.x=2*temp.x;
            if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)
            {
                visit[temp1.x]=true;
                lqueue.push(temp1);
            }
        }
        while(!lqueue.empty())
        {
            lqueue.pop();
        }
        cout<<min_time<<endl;
    }

    return 0;
}