Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 210291 Accepted: 63838
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.
题目大意
给一个起始点 N 和终点 K , 假设当前位置为 X , 有三种移动方式，第一种是往左走一个单位长度，即变成 X - 1 ; 第二种是往右走一个单位长度，即变成 X + 1 ; 第三种是从 X 变成 2 * X 。 每种移动方式消耗的时间都是一分钟。求得从起始点到终点的最短耗费时间。
解题思路
这道题很简单，就是在一个X轴上的BFS求最短路模型，只要按照思路写就行。
/* 一切都是命运石之门的选择 El Psy Kongroo */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<cmath>
#include<functional>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<int, pii> piii;
typedef pair<double, double> pdd;
typedef pair<string, pii> psi;
typedef __int128 int128;
#define x first
#define y second
const int inf = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 1e5 + 10;
int sx, tx; //起点 终点
int dist[N];
//判断位置是否合法
bool check(int x){
return x >= 0 && x < N && dist[x] == -1;
}
int bfs(){
memset(dist, -1, sizeof(dist));
queue<int> q; q.push(sx);
dist[sx] = 0;
while(!q.empty()){
int x = q.front();
q.pop();
if(x == tx) return dist[x]; //到达终点
if(check(x - 1)) q.push(x - 1), dist[x - 1] = dist[x] + 1;
if(check(x + 1)) q.push(x + 1), dist[x + 1] = dist[x] + 1;
if(check(2 * x)) q.push(2 * x), dist[2 * x] = dist[x] + 1;
}
return -1;
}
int main(){
cin >> sx >> tx;
int res = bfs();
cout << res << endl;
return 0;
}