Catch That Cow

Problem 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.

题目描述:找最短步数,用bfs

第一种:含标记数组

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int n, k,ans,step[200010],book[200010]; //wa了好几次,这里一定要大一点 ,step:步数 book:标记走没走
void bfs(int a, int b) {
ans = 0;
book[a] = 1;
queue<int >q;
q.push(a);
while (!q.empty()) {
int x = q.front();
//cout <<x<<" "<<b<<"\n";
q.pop();
if( x== b)break; if (x * 2 <= 200010 && x * 2 >= 0 && book[x * 2] == 0) { //判断边界,判断走没走,
book[2 * x] = 1;
q.push(x * 2);
step[x * 2] = step[x] + 1;
}
if (x + 1 <= 200010 && x + 1 >= 0 && book[x + 1] == 0) { //判断边界,判断走没走,
book[1 + x] = 1;
q.push(x +1);
step[x + 1] = step[x] + 1;
}
if (x - 1 <= 200010 && x - 1 >= 0 && book[x - 1] == 0) { //判断边界,判断走没走,
book[x - 1] = 1;
q.push(x -1);
step[x - 1] = step[x] + 1;
}
}
}
int main() {
while (cin >> n >> k) {
memset(step, 0, sizeof(step)); //不要忘记初始化
memset(book, 0, sizeof(book));
if (n >= k)cout << n - k << endl; //n不大于k的话 就是n-k了
else {
bfs(n, k);
cout << step[k] << endl;
}
}
return 0;
}

第二种:

不加标记数组,思考了两天了,我一直认为可以不加标记数组,但是提交就wa了,找了两天原因没找到,终于发现了,是我判断的顺序不对,一定要最后判读2*x,否则先判断2*i,后边会越来越大,所以你就不好把握最大值了,并且容易超时,所以先判断x-1  

判断顺序很重要

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int n, k,step[200010];
int bfs(int a, int b) {
queue<int >q; //初始化
memset(step, 0, sizeof(step));
q.push(a); while (!q.empty()) {
int x = q.front();
q.pop(); if (x == b)break; //找到就终止 if (x - 1 <= 200010 && x - 1 >= 0 && step[x -1] == 0) { //这三个顺序很重要,
q.push(x - 1);
step[x - 1] = step[x] + 1;
}
if (x + 1 <= 200010 && x + 1 >= 0 && step[x +1] == 0) {
q.push(x + 1);
step[x + 1] = step[x] + 1;
}
if (x * 2 <= 200010 && x * 2 >= 0 && step[x * 2]==0) { //这个一定要放在最后边
q.push(x * 2);
step[x * 2] = step[x] + 1;
}
}
return step[b];
}
int main() {
while (cin >> n >> k) { if (n >= k) //因为n减小只能-1,所以直接输出就可以
cout << n - k << endl;
else
cout << bfs(n, k)<< endl;
}
return 0;
}
 

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