标签：any name return == ace tran lang code form
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
Output
Sample Input
5 17
Sample Output
4
Hint
#include <algorithm> #include <stdio.h> #include <queue> #include <string.h> #include <stdlib.h> #include <iostream> using namespace std; const int N = 1000000; bool book[1000010]; int n, k; struct node{ int num; int step; }; bool check(int x){ if(x<0 || x>N || book[x]) return false; return true; } void BFS(){ node q; q.num = n; q.step = 0; queue<node> qq; qq.push(q); book[n] = 1; while(!qq.empty()){ node t = qq.front(); qq.pop(); if(t.num==k){ cout<<t.step<<endl; return; } node tt = t; tt.num = t.num*2; if(check(tt.num)){ book[tt.num] = true; tt.step = t.step+1; qq.push(tt); } tt.num = t.num-1; if(check(tt.num)){ book[tt.num] = true; tt.step = t.step+1; qq.push(tt); } tt.num = t.num+1; if(check(tt.num)){ book[tt.num] = true; tt.step = t.step+1; qq.push(tt); } } printf("-1\n"); return; } int main() { while(~scanf("%d%d",&n,&k)){ memset(book,false,sizeof(book)); BFS(); } return 0; }
标签：any name return == ace tran lang code form
原文地址：https://www.cnblogs.com/0526yao/p/10508151.html
