poj3278-catch that cow(bfs广搜)

先贴题目
Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 59759 Accepted: 18581
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

题意就是从一个数经过+1，-1， *2 如何用最小的操作次数达到另一个数。

找最短路径和最优解就是广搜没跑了。

#include<iostream>#include<queue>#include<cstring>#define INF 10e7const int M = 100001;using namespace std;int dir[3][2] = {{1,1}, {1,-1}, {2,0}};struct POINT{    int pos,        step;}now, next;queue<POINT> Q;bool visit[M];int N, K;int bfs(){    while(!Q.empty())        Q.pop();    now.pos = N;    now.step = 0;    visit[now.pos] = true;    Q.push(now);    while(!Q.empty())    {        now = Q.front();        Q.pop();        next = now;        for(int i = 0; i < 3; i ++)        {            next.pos = now.pos * dir[i][0] + dir[i][1];            next.step = now.step + 1;            if(next.pos < 0 || next.pos > M)                continue;            else if(next.pos == K)                return next.step;            else if(!visit[next.pos])            {                Q.push(next);                visit[next.pos] = true;            }        }    }    return INF;}int main(){    while(cin >> N >> K)    {        memset(visit,false,sizeof(visit));        if(N < K)            cout << bfs() << endl;        if(N == K)            cout << 0 << endl;        if(N > K)            cout << N - K << endl;    }}

还是经典的写法， 不过这个函数没有用参数，大概是因为都使用了全局变量吧。刚学的时候写起来还是很难的，看都看不懂，多写几次就会好一点。