搜索C Catch That Cow POJ - 3278

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zhaohaibo_/article/details/85010023

写在开头：
可能是我对题目的理解有误，但我认为本题的测试数据实在是坑：

1. 测试数据有多组，而不是单组
2. 数据范围实际超过100000，数组需要开到500000

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.

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <map>
#define INF 0x3f3f3f3f
using namespace std;

struct node{
    int pos;
    int step;
    node(){}
    node(int pos,int step):pos(pos),step(step){}
};
const int maxn = 500050;
int book[maxn];
int k,n;
int Min = INF;
int bfs(){
    node head, tail;
    queue<node>que;
    que.push(node(n, 0));
    book[n] = 1;
    while(!que.empty()){
        head = que.front();
        que.pop();
        if(!book[head.pos + 1] && head.pos < 100000){
            if(head.pos + 1 == k) return head.step + 1;
            book[head.pos + 1] = 1;
            que.push(node(head.pos + 1, head.step + 1));
        }
        if(!book[head.pos - 1] && head.pos > 0){
            if(head.pos - 1 == k) return head.step + 1;
            book[head.pos - 1] = 1;
            que.push(node(head.pos - 1, head.step + 1));
        }
        if(!book[head.pos << 1] && head.pos < 100000){
            if(head.pos << 1 == k) return head.step + 1;
            book[head.pos << 1] = 1;
            que.push(node(head.pos << 1, head.step + 1));
        }
    }
    return 0;
}
int main(){
    ios::sync_with_stdio(false);
    while(cin >> n >> k){
        memset(book, 0, sizeof book);
        cout << bfs() << endl;
    }
    return 0;
}