Poj3278 Catch That Cow ( BFS

最新推荐文章于 2024-07-24 11:36:30 发布

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最新推荐文章于 2024-07-24 11:36:30 发布

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本文详细解析了CatchThatCow问题的算法实现，该问题描述了农夫约翰如何通过步行和传送的方式，在最短时间内找到静止不动的逃逸奶牛。文章提供了一段使用广度优先搜索（BFS）算法的C++代码示例，并解释了如何根据起始位置和目标位置计算最短路径。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 88609		Accepted: 27757

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

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define ll long long
#define N 200010           
const int mod = 1e9+7;
int n, k;
struct node {
    int x, step; 
}p[N];                  //*2数据要开大两倍2233 没开会RE 
queue<node> q; 
bool vis[N];
int bfs()
{
    vis[n] = true;
    p[n].x = n;
    p[n].step = 0;
    q.push(p[n]);
    int pos, step;
    while(!q.empty()) {
        pos = q.front().x;
        step = q.front().step;
        q.pop();
        if(pos==k) return step;
        if(!vis[pos-1] && pos>0) {   // pos必须大于0 
            vis[pos-1] = true;
            p[pos-1].x = pos-1;
            p[pos-1].step = step+1;
            q.push(p[pos-1]);
        }
        if(pos<k) {
            if(!vis[pos+1]) {
            vis[pos+1] = true;
            p[pos+1].x = pos+1;
            p[pos+1].step = step+1;
            q.push(p[pos+1]);
            }
            if(!vis[pos*2]) {
            vis[pos*2] = true;
            p[pos*2].x = pos*2;
            p[pos*2].step = step+1;
            q.push(p[pos*2]);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&k)) {
        memset(vis,false,sizeof(vis));
        memset(p,0,sizeof(p));
        while(!q.empty()) q.pop();
        printf("%d\n",bfs());

    }
return 0;
}