poj3278Catch That Cow(BFS)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37094		Accepted: 11466

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.

刚做的时候发生了很多奇异YY, 搞的我不知道重写了多少遍。。

从当前位置local，向local+1, local-1, local*2这三个方向扩展。

code：

#include <queue>

#include <cstdio>

#include<cstring>

#define N 100001

using namespace std;

int n, k, ans;

bool vis[N];

struct node {

    int local, time;

};

void bfs() {

    int t, i;

    node now, tmp;

    queue<node> q;

    now.local = n;

    now.time = 0;

    q.push(now);

    memset(vis,false,sizeof(vis));

    vis[now.local] = true;

    while(!q.empty()) {

        now = q.front();

        q.pop();

        for(i=0; i<3; i++) {

            if(0==i) t = now.local-1;

            else if(1==i) t = now.local+1;

            else if(2==i) t = now.local*2;

            if(t<0||t>N||vis[t]) continue;

            if(t==k) {

                ans = now.time+1;

                return;

            }

            vis[t] = true;

            tmp.local = t;

            tmp.time = now.time+1;

            q.push(tmp);

        }

    }

}



int main() {

    while(~scanf("%d%d",&n,&k)) {

        if(n>=k) printf("%d\n",n-k);

        else {

            bfs();

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

        }

    }

}