POJ 3278 Catch That Cow(搜索BFS)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27816		Accepted: 8556

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

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

解题报告：

这道题其实挺简单的，但是好久没有写搜索题了，很不细心，

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1000010;
int visit[2 *N], queue[2 * N], step[2 * N], ans;//queue起到队列的作用
void BFS(int begin, int end)
{
    int rear, front, a, b[3], i;
    rear = 0;//队列的尾
    front = 0;//队列的头
    rear ++;
    queue[rear] = begin;//初始化
    while (front < rear)
    {
        front ++;
        a = queue[front];
        if (a == end)//找到
        {
            ans = step[a];
            return ;
        }
        else
        {
            b[0] = a + 1;
            b[1] = a - 1;
            b[2] = 2 * a;
            for (i = 0; i <=2; ++i)
            {
                if (b[i] >= 0 && b[i] <= 1000000 && !visit[b[i]])//当时写的时候1000000写成了100000少写了一个零，Runtime Error好几次！！
                {
                    visit[b[i]] = 1;//标记
                    rear ++;
                    queue[rear] = b[i];//插到队尾
                    step[b[i]] = step[a] + 1;//步数加1
                }
            }
        }
    }
}
int main()
{
    int  begin, end;
    while (scanf("%d%d", &begin, &end) != EOF)
    {
        memset(visit, 0, sizeof(visit));
        memset(queue, 0, sizeof(queue));
        memset(step, 0, sizeof(step));
        visit[begin] = 1;
        BFS(begin, end);
        printf("%d\n", ans);
    }
    return 0;
}

，哎！这道题的题意：就是在一条数轴上，求起始点到终点的最小步数，很显然得用BFS借助于队列；

代码如下：