POJ 3278解题报告(C语言版)//Catch That Cow
来源:互联网 发布:淘宝代理靠不靠谱 编辑:程序博客网 时间:2024/06/09 19:14
广搜的简单运用
B - Catch That Cow
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
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<stdio.h>#include<string.h>struct A{ int s;//位置 int point;//步数}queue[1000000];//可以将其看成一个数组int sign[1000000];//标记是否走过,1为已走,0为未走int bfs(int n,int k){ int head=0,rear=0; struct A r; sign[n]=1;//农夫开始位置做标记 queue[rear].s=n,queue[rear].point=0,rear++;//入队 while(head<rear) { r=queue[head++];//出队 if(r.s+1>=0&&r.s+1<=100000&&!sign[r.s+1]) { queue[rear].point=r.point+1; queue[rear++].s=r.s+1; sign[r.s+1]=1; if(queue[rear-1].s==k) return queue[rear-1].point; } if(r.s-1>=0&&r.s-1<=100000&&!sign[r.s-1]) { queue[rear].point=r.point+1; queue[rear++].s=r.s-1; sign[r.s-1]=1; if(queue[rear-1].s==k) return queue[rear-1].point; } if(r.s*2>=0&&r.s*2<=100000&&!sign[r.s*2]) { queue[rear].point=r.point+1; queue[rear++].s=r.s*2; sign[r.s*2]=1; if(queue[rear-1].s==k) return queue[rear-1].point; } } return 0;}int main(){ int n,k; while(scanf("%d%d",&n,&k)!=EOF) { memset(sign,0,sizeof(sign)); printf("%d\n", bfs(n,k)); } return 0;}
- POJ 3278解题报告(C语言版)//Catch That Cow
- POJ 3278 Catch That Cow【C语言版】
- POJ-3278 Catch That Cow 解题报告
- POJ - 3278 Catch That Cow解题报告
- Catch That Cow解题报告
- POJ 3278(Catch That Cow)解题纠错
- POJ3278 Catch That Cow ACM解题报告(BFS)
- POJ_3278 Catch That Cow 解题报告
- 【poj 3278】Catch That Cow 题意&题解&代码(C++)
- poj 3278 Catch That Cow(bfs)
- POJ 3278 Catch That Cow (BFS)
- POJ 3278 Catch That Cow(BFS)
- POJ--3278:Catch That Cow (BFS)
- POJ 3278 - Catch That Cow(BFS)
- POJ 3278 Catch That Cow (BFS)
- POJ - 3278 - Catch That Cow (BFS)
- POJ 3278 Catch That Cow (BFS)
- POJ 3278 Catch That Cow (BFS)
- Qt核心剖析:信息隐藏(1)
- sql语句
- Tuple的取值过程/求长度
- js document常用的属性以及方法
- UVa 10036 Divisibility (同余DP)
- POJ 3278解题报告(C语言版)//Catch That Cow
- Java:单例模式的七种写法
- Unity3D基础学习之AssetBundle 资源包创建与加载
- Android 关机解析
- js根据15,18位身份证号获取生日与性别
- 爱一个人就要不顾一切
- 通过标签用法对比HTML 5开发与HTML 4的区别
- 学习jQuery必须知道的几种常用方法
- Eclipse报错:Setting property 'source' to 'org.eclipse.jst.jee.server:test1' did no