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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
Output
Sample Input
5 17
Sample Output
4
Hint
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<queue> #include<algorithm> using namespace std; #define INF 0xfffffff #define N 100010 int m,n; int vis[N]; struct node { int x,step; friend bool operator<(node a,node b) { return a.step>b.step; } }; int dfs() { priority_queue<node>Q; memset(vis,0,sizeof(vis)); node q,s; s.x=n; vis[s.x]=1; s.step=0; Q.push(s); int i; while(!Q.empty()) { q=Q.top(); Q.pop(); if(q.x==m) return q.step; for(i=0;i<3;i++) { if(i==0) s.x=q.x+1; else if(i==1) s.x=q.x-1; else if(i==2) s.x=q.x*2; if(s.x<100001&&s.x>=0&&vis[s.x]==0)//vis[s.x]==0必须放到后面，-_-被运行错误错了好多次； { vis[s.x]=1; s.step=q.step+1; Q.push(s); } } } return -1;//要有返回值，我也不知道为什么； } int main() { while(scanf("%d%d",&n,&m)!=EOF) { if(m==n) { printf("0\n"); continue; } int ans; ans=dfs(); printf("%d\n",ans); } return 0; }
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原文地址：http://www.cnblogs.com/zhengguiping--9876/p/4471908.html
