The title is to give you the starting point sx and the ending point gx; the cow can perform the following two operations at the starting point:
Walk: John spends one minute moving from any point X to point X-1 or point X+1.
Teleport: John spends one minute moving from any point X to point 2*X.
You want to output the shortest steps and print the path.
The shortest number of steps is to use bfs.
As for the path, you can use a structure to store the parent node of each point, and then recursively output the path.
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
#define MAX 100010
int sx,gx,ans;
struct mask
{
mask(int x,int step):x(x),step(step){};
int x,step,f;
};/
struct node
{
int x;
}p[MAX];//保存父节点
queue<mask>q;
int dx[]={1,-1};
int flag=0;
bool vis[MAX];
bool check(int r)
{
return (0<=r&&r<=MAX);
}
void bfs()
{
flag=0;
while(!q.empty())q.pop();
q.push(mask(sx,0));
memset(vis,false,sizeof(vis));
vis[sx]=true;//printf("%d\n",q.front());
while(q.size())
{
mask tmp=q.front();q.pop();
// printf("ok\n");
if(tmp.x==gx)
{
ans=tmp.step;
flag=1;
break;
}
for(int i=0;i<2;i++)
{
int nx=tmp.x+dx[i];
if(check(nx)&&!vis[nx])
{
vis[nx]=true;
p[nx].x=tmp.x;
q.push(mask(nx,tmp.step+1));
}
}
int nx1=tmp.x*2;
if(check(nx1)&&!vis[nx1])
{
vis[nx1]=true;
p[nx1].x=tmp.x;
q.push(mask(nx1,tmp.step+1));
}
}
}
void pri(int x1)
{
if(x1==sx){
printf("%d ",sx);
return ;
}
pri(p[x1].x);
printf("%d ",x1);
}
int main()
{
while(~scanf("%d %d",&sx,&gx)){
if(sx==-1&&gx==-1)break;
bfs();
if(flag)
printf("%d\n",ans);
else printf("-1\n");
pri(gx);
printf("\n");
}
}