POJ3278：Catch That Cow（BFS）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 82031		Accepted: 25785

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

用BFS即可，有地方需要注意，①visit数组要开到200000以上，否则“心灵传输”时可能会下标越界，②当上一个状态坐标超过K时不用向前走或“心灵传输”了，防止越界和优化时间。

# include <stdio.h>
# include <queue>
# include <string.h>
using namespace std;
struct node
{
    int x, step;
};
queue<node>Q;
int vis[200001];
int bfs(int pos, int target)
{
    node tmp;
    tmp.x = pos;
    tmp.step = 0;
    Q.push(tmp);
    while(!Q.empty())
    {
        node tmp = Q.front();
        if(tmp.x == target)
            return tmp.step;
        if(tmp.x > 0 && !vis[tmp.x-1])
        {
            node temp;
            vis[tmp.x-1] = 1;
            temp.x = tmp.x-1;
            temp.step = tmp.step + 1;
            Q.push(temp);
        }
        if(tmp.x <= target && !vis[tmp.x+1])
        {
            node temp;
            vis[tmp.x+1] = 1;
            temp.x = tmp.x+1;
            temp.step = tmp.step + 1;
            Q.push(temp);
        }
        if(tmp.x <= target && !vis[(tmp.x)<<1])
        {
            node temp;
            vis[(tmp.x)<<1] = 1;
            temp.x = (tmp.x)<<1;
            temp.step = tmp.step + 1;
            Q.push(temp);
        }
        Q.pop();
    }
}
int main()
{
    int n, m;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        if(m <= n)
        {
            printf("%d\n",n-m);
            continue;
        }
        memset(vis, 0, sizeof(vis));
        while(!Q.empty())
            Q.pop();
        printf("%d\n",bfs(n, m));
    }
    return 0;
}