Catch That Cow(poj 3278)
给定两个整数n和k
通过 n+1或n-1 或n*2 这3种操作,使得n==k
输出最少的操作次数
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
//广搜,a是队列,step记录步数,vis记录哪些数被搜到过 #include<cstdio> #include<iostream> #define M 1000010 using namespace std; int step[M],a[M],vis[M]; int main() { int n,m,head=0,tail=1; scanf("%d%d",&n,&m); a[1]=n; vis[n]=1; while(head<tail) { ++head; int u=a[head]; if(u==m) { printf("%d",step[head]); return 0; } if(u-1>=0&&!vis[u-1]) { a[++tail]=u-1; step[tail]=step[head]+1; vis[u-1]=1; } if(u+1<1000000&&!vis[u+1]) { a[++tail]=u+1; step[tail]=step[head]+1; vis[u+1]=1; } if(u*2<1000000&&!vis[u*2]) { a[++tail]=u*2; step[tail]=step[head]+1; vis[u*2]=1; } } return 0; }