标签：des style blog class c code
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 43674 | Accepted: 13612 |
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
Source
题目大意：
告诉你起点位置和终点位置，你可以有三种操作，从某个位置x，可以到达x-1,x+1,2*x，问你从起点到终点最短时间？
解题思路：
利用bfs枚举位置即可
但是刨除以下情况，也就是剪枝一下
1、当x<=0 时，x-1不能到达，这个状态剪掉
2、当x>终点位置时，x+1和2*x均不能到达
解题代码：
#include <iostream>
#include <queue>
using namespace std;
const int maxn=210000;
int marked,visited[maxn];
struct node{
int pos,t;
node(int pos0=0,int t0=0){
pos=pos0;t=t0;
}
};
int main(){
int from,to;
while(cin>>from>>to){
marked++;
queue <node> q;
q.push(node(from,0));
visited[from]=marked;
while(!q.empty()){
node s=q.front();
q.pop();
if(s.pos==to){
cout<<s.t<<endl;
break;
}
if( s.pos-1>=0 && visited[s.pos-1]!=marked){
visited[s.pos-1]=marked;
q.push(node(s.pos-1,s.t+1));
}
if( s.pos<=to && visited[s.pos+1]!=marked){
visited[s.pos+1]=marked;
q.push(node(s.pos+1,s.t+1));
}
if( s.pos<=to && visited[s.pos*2]!=marked){
visited[s.pos*2]=marked;
q.push(node(s.pos*2,s.t+1));
}
}
}
return 0;
}
POJ 3278 Catch That Cow,布布扣,bubuko.com
标签：des style blog class c code
原文地址：http://blog.csdn.net/a1061747415/article/details/26343741
