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.
#include<iostream> #include<cstring> #include<string> #include<cmath> #include<map> #include<queue> #include<cstdio> #include<vector> #include<algorithm> #define bug printf("-----\n"); using namespace std; const int maxn=100011; const int inf=210000; typedef long long ll; int d[maxn]; bool vis[maxn]; //从当前的X出发,可以到达X+1,X-1,X*2,三个位置处,明显X-1不能成为负数,x+1 也不能大于K,如果大于了就会增加步数, //但是对于X*2这个条件,可以在保证X<K的情况下,计算X*2的值,而不是保证X*2的值小于K void bfs(int x,int k) { queue<int> q; memset(vis,false,sizeof(vis)); memset(d,0,sizeof(d)); q.push(x); vis[x]=true; d[x]=0; while(!q.empty()) { int x=q.front(); q.pop(); if(x==k)return ; if(x-1>=0&&!vis[x-1]) { vis[x-1]=true; q.push(x-1); d[x-1]=d[x]+1; } if(x+1<=k&&!vis[x+1]) { vis[x+1]=true; q.push(x+1); d[x+1]=d[x]+1; } if(x<=k&&x*2<=100000&&!vis[x*2]) { vis[x*2]=true; q.push(x*2); d[x*2]=d[x]+1; } } } int main() { int n,m,i,j,t,k; //freopen("in.txt","r",stdin); while(~scanf("%d%d",&n,&k)) { bfs(n,k); printf("%d\n",d[k]); } return 0; }