Catch That Cow
题目描述：
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?
输入：
Line 1: Two space-separated integers: N and K
输出：
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
提示：
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.
题目的意思是，知道农夫和牛的位置，农夫有两种移动方式——移动一格或是传送到当前位置坐标的两倍处，求抓到牛的最短步数。
解题思路启发：https://blog.csdn.net/qq_38620461/article/details/78445374
这题是我做的第一道，也是一道入门的BFS，大概的解题思路就是：从起点开始，一层一层向下遍历，直到遍历出正确答案，需要用到结构体来记录位置和当前的深度，用queue来一层层遍历。
这是第一发代码：
#include#include #include using namespace std; int vis[100005]; struct point { int place,count; }; int main() { int n,k; point x,temp; while(~scanf("%d %d",&n,&k)) { if(n==k) {printf("%d\n",0);continue;} fill(vis,vis+100005,0); queue q; point head={n,0}; q.push(head); vis[n]=1; while(!q.empty()) { x=q.front(); q.pop(); temp=x; temp.place-=1; if(temp.place>=0&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp); } temp=x; temp.place+=1; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp); } temp=x; temp.place*=2; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp); } } } return 0; }
然后，就碰到了我人生中的第一个Memory Limit Exceeded...
很快意识到漏掉了对搜索过的位置进行标记的步骤，于是作出更改，顺手把标记数组改成bool型，节省空间，之后
信心满满的第二发：
#include#include<string.h> #include #include using namespace std; bool vis[100005]; struct point { int place,count; }; int main() { int n,k; point x,temp; while(~scanf("%d %d",&n,&k)) { if(n==k) {printf("%d\n",0);continue;} memset(vis,0,sizeof(vis)); queue q; point head={n,0}; q.push(head); vis[n]=1; while(!q.empty()) { x=q.front(); q.pop(); temp=x; temp.place-=1; if(temp.place>=0&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]==1; } temp=x; temp.place+=1; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]==1; } temp=x; temp.place*=2; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]==1; } } } return 0; }
然而，依旧是Memory Limit Exceeded...
自己又跑了一遍，输入了比较大的数字，电脑的内存瞬间满了
这下就被迷住了，反复比对了正确代码和自己的代码，感觉整体上差不多一样啊。。。
最后经过了2个小时的debug，终于发现是将赋值运算符打成了关系运算符，改正之后，终于ac了= =
第三发：
#include#include<string.h> #include #include using namespace std; bool vis[100005]; struct point { int place,count; }; int main() { int n,k; point x,temp; while(~scanf("%d %d",&n,&k)) { if(n==k) {printf("%d\n",0);continue;} memset(vis,0,sizeof(vis)); queue q; point head={n,0}; q.push(head); vis[n]=1; while(!q.empty()) { x=q.front(); q.pop(); temp=x; temp.place-=1; if(temp.place>=0&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]=1; } temp=x; temp.place+=1; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]=1; } temp=x; temp.place*=2; if(temp.place<=100000&&vis[temp.place]==0) { ++temp.count; if(temp.place==k) {printf("%d\n",temp.count);break;} q.push(temp);vis[temp.place]=1; } } } return 0; }
不过我还是不知道为什么关系运算符会让它爆内存
纪念第一个内存超限