题目链接:http://poj.org/problem?id=3278

Catch That Cow
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 124528   Accepted: 38768

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 - 1 or + 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.

题意:给你两个点 N,K,你有三种走法:X+1 ,X-1,X*2 ,求从N到K走的最少步数。

代码附有Wrong Answer 和 Runtime Error的几个参考原因。

 #include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std; /**
* 造成Runtime Error的原因有两点:
* 1.数组开小了,也就是下面的maxn可能只开到 1e5稍多一点,这是不对的,在搜索过程中有个 2*x ,这会导致数组大小应该大于1e5的两倍。
* 2.在判断!mark[x-1]的时候,没有x>0这一个约束条件,如果x = 0 ,则 x-1会导致数组越界;
* 3.在 x+1 和 2*x这两种情况下,应该限制 x<=K;
*/
/**
* 造成Wrong Answer 的原因之一:在判断 x-1 的情况的时候,不要有 x<=K
*/
const int maxn = 2e5+;
bool mark[maxn];
int N,K;
struct node
{
int x;
int step;
}; //队列的写法;
void bfs()
{
queue<node> q;
struct node cu,ne;
cu.x = N;
cu.step = ;
mark[N] = ;
q.push(cu);
while(!q.empty())
{
cu = q.front();
q.pop();
if(cu.x == K)
{
printf("%d\n",cu.step);
return ;
}
//下面的三种情况应该是并列关系,不应该满足一种情况就不判断别情况了
//我开始写成了if-else if-else if形式,想成不是并列的形式了;
//如果实在找不到哪里错了,可以打印cu.x变量,根据变量的值找问题;
if(cu.x>&&!mark[cu.x-])
{
ne.x = cu.x - ;
ne.step = cu.step + ;
mark[ne.x] = ;
q.push(ne);
}
if(cu.x<=K&&!mark[cu.x+])
{
ne.x = cu.x + ;
ne.step = cu.step + ;
mark[ne.x] = ;
q.push(ne);
}
if(cu.x<=K&&!mark[cu.x*])
{
ne.x = *cu.x;
ne.step = cu.step + ;
mark[ne.x] = ;
q.push(ne);
}
}
}
int main()
{
while(scanf("%d%d",&N,&K)!=EOF)
{
memset(mark,,sizeof(mark));
bfs();
}
return ;
}

Ac代码

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