POJ(3278)：Catch That Cow

ime Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 119162		Accepted: 37180

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

解题思路：

DFS,BFS都可以都可以，就这道题而言，深搜比较慢。

按层便利每个元素，找到牛的位子就输出深度。

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 100000;
bool visited[maxn+10];
struct step{
    int x;
    int steps;
    step(int xx,int y):x(xx),steps(y) {}
};

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    queue<step> que;
    que.push(step(n,0));
    visited[n] = 1;
    memset(visited,0,sizeof(visited));
    while(!que.empty()){
        step s = que.front();
        if(s.x == k){
            printf("%d",s.steps);
            return 0;
        }
        else{
            if(s.x - 1 >= 0 && !visited[s.x - 1]){
                que.push(step(s.x-1,s.steps+1));
                visited[s.x - 1] = 1;
            }
            if(s.x + 1 <= maxn && !visited[s.x + 1]){
                que.push(step(s.x+1,s.steps+1));
                visited[s.x + 1] = 1;
            }
            if(2*s.x <= maxn && !visited[2 * s.x]){
                que.push(step(2*s.x,s.steps+1));
                visited[2 * s.x] = 1;
            }
        }
        que.pop();
    }
    return 0;
}