POJ 3278 Catch That Cow (BFS)

Catch That CowTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 49787 Accepted: 15623DescriptionFarmer John has been informed of the location of a fugitive

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590人浏览 · 2015-02-10 09:35:05

_TCgogogo_ · 2015-02-10 09:35:05 发布

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 49787		Accepted: 15623

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤N ≤ 100,000) on a number line and the cow is at a pointK (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 pointsX- 1 orX+ 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 andK

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

题目链接：http://poj.org/problem?id=3278

题目大意：一条线上，给一个起点和一个终点，有三种走法，走一次1秒，位置坐标加1，减1和乘2，求从起点到终点的最短时间

题目分析：裸BFS，有可能n >= k，这时候只需要求n-k即可，n < k时，搜一搜就行了，千万注意！！虽然肯定有解，但是在BFS函数最后还是要return一个值，不然会wa

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int n, k;
bool vis[200005];

struct Node
{
    int num;
    int step;
};

int BFS()
{
    memset(vis, false, sizeof(vis));
    queue <Node> q;
    Node st, cur, t;
    st.num = n;
    st.step = 0;
    vis[n] = true;
    q.push(st);
    while(!q.empty())
    {
        t = q.front();
        cur = t;
        q.pop();
        if(t.num == k)
            return t.step;
        if(t.num + 1 < 200001 && !vis[t.num + 1])
        {
            vis[t.num + 1] = true;
            t.num ++;
            t.step ++;
            q.push(t);
        }
        t = cur;
        if(t.num - 1 >= 0 && !vis[t.num - 1])
        {
            vis[t.num - 1] = true;
            t.num --;
            t.step ++;
            q.push(t);
        }
        t = cur;
        if(t.num * 2 < 200001 && !vis[t.num * 2])
        {
            vis[t.num * 2] = true;
            t.num *= 2;
            t.step ++;
            q.push(t);
        }
    }
    return 0;
}

int main()
{
    while(scanf("%d %d", &n, &k) != EOF)
    {
        if(n >= k)
            printf("%d\n", n - k);
        else
            printf("%d\n", BFS());
    }
}