Catch That Cow解题报告

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我是靠谱客的博主无语刺猬，最近开发中收集的这篇文章主要介绍Catch That Cow解题报告，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

B - Catch That Cow

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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

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.

这道题目突出一个心酸啊！！

通过BFS进行解决方案的搜索。用一个队列来储存中间数。

但是这个队列的大小很难把握，我开始以为10000算大了，显示RE错误，后来我对储存进行优化。

凡是访问过的统统不再加入队列中，已经在队列中的也不再加入队列，比结果还要大1以上的也不加入队列，果然大大减少了队列的数据，但是还是RUNTIME ERROR错误。

然后我改为100000，还是RE，最后一狠心改为1000000才不显示RE，反倒显示了Wrong Answer。我反复查看代码似乎没有问题。

原来我疏忽了一种N比K还要大的可能性。一旦N比K大的话，只有减一才能逐步到达K，所以步数就是N - K。

这是我的源代码：

//Catch That Cow
#include <stdio.h>
#define SIZE 1000000
long num[SIZE] = {0};
long step[SIZE] = {0};
int visited[SIZE] = {0};
long i,j;
void add(long n);
void main()
{
long n,k;
scanf("%d %d",&n,&k);
num[0] = n;
i = 0;
j = 1;
if (n <= k)
{
while (num[i % SIZE] != k)
{
visited[num[i % SIZE]] = 1;
if (!(num[i % SIZE] > k + 1 || num[i % SIZE] < 0))
{
if (!visited[num[i % SIZE] + 1])
{
add(num[i % SIZE] + 1);
}
if (!visited[num[i % SIZE] - 1])
{
add(num[i % SIZE] - 1);
}
if (!visited[num[i % SIZE] * 2])
{
add(num[i % SIZE] * 2);
}
}
i ++;
}
printf("%dn",step[i % SIZE]);
}
else
{
printf("%dn",n - k);
}
}
void add(long n)
{
num[j % SIZE] = n;
visited[n] = 1;
step[j % SIZE] = step[i % SIZE] + 1;
//printf("add %d step %dn",n,step[j % SIZE]);
j ++;
}