POJ_3278 Catch That Cow 解题报告
2012-03-16 01:21
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Catch That Cow
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
Sample Output
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.
题目连接:http://poj.org/problem?id=3278
算法类型:BFS
解题思路:BFS搜索,标记已经访问的进行剪枝。
算法实现:
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 28177 | Accepted: 8677 |
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.
题目连接:http://poj.org/problem?id=3278
算法类型:BFS
解题思路:BFS搜索,标记已经访问的进行剪枝。
算法实现:
#include<stdio.h> #include<string.h> #include<stdlib.h> int vis[100000]; //记录是否访问过; int dis[100000]; //记录步数; int q[100000]; //队列; int n,k; int bfs(int n) //广搜函数; { int rear=0 ,front=0,u; //队列头和尾都为0; vis ==1; q[rear++]=n; //入队列; while(front<rear) //队列不为空; { u=q[front++]; //出队列; if(u==k) // 当找到牛时就跳出; break; if(u*2<=100000 && vis[2*u]==0 && u*2>0) //下面三个IF 是广搜的三个方向; { dis[u*2]=dis[u]+1; //记录达到这点走过的步数; q[rear++]=u*2; //入队列; vis[u*2]=1; } if(u+1<=100000 && vis[u+1]==0 && u+1>0) { dis[u+1]=dis[u]+1; q[rear++]=u+1; vis[u+1]=1; } if(u-1<=100000 && u-1>0 && vis[u-1]==0) { dis[u-1]=dis[u]+1; q[rear++]=u-1; vis[u-1]=1; } } return 1; } int main() { while(scanf("%d%d",&n,&k)!=EOF) { memset(vis,0,sizeof(vis)); //置0函数; memset(q,0,sizeof(q)); memset(dis,0,sizeof(dis)); if(k<=n) //判断是人在前,还是牛在前; { printf("%d\n",n-k); // 输出步数; } else { bfs(n); //调用宽搜函数; printf("%d\n",dis[k]); // 输出步数; } } return 0; }
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