bzoj1646[Usaco2007 Open]Catch That Cow 抓住那只牛*

bzoj1646[Usaco2007 Open]Catch That Cow 抓住那只牛

题意:

数轴上,起点在n,终点在k,每次走可以向左走一步或向右走一步或瞬移到当前坐标的两倍位置,问最少走几次。0≤n,k≤100000。

题解:

bfs,允许走的位置边界为[0,max(n,k)+1]。下界为0原因是如果走到小于0的位置,k≥0,则瞬移和往左走都是南辕北辙,只能向右走,那么一开始就不应该走到小于0的位置导致浪费时间。上界为max(n,k)+1的原因是如果你走到了大于这个数的位置,k必定小于当前位置,则你必须一步一步的往回走,而这样做显然比之前就别走到这个位置花更多时间……(我承认我乱扯一通)所以复杂度为100000。

代码:

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <queue>
 5 #define maxn 100100
 6 #define inc(i,j,k) for(int i=j;i<=k;i++)
 7 using namespace std;
 8
 9 queue <int> q; int tim[maxn],n,k,mx;
10 int main(){
11     scanf("%d%d",&n,&k); memset(tim,-1,sizeof(tim)); q.push(n); tim[n]=0; mx=max(n,k)+1;
12     while(!q.empty()){
13         int x=q.front(); q.pop();
14         if(x+1<=mx&&tim[x+1]==-1){q.push(x+1); tim[x+1]=tim[x]+1; if(x+1==k)break;}
15         if(x-1>=0&&tim[x-1]==-1){q.push(x-1); tim[x-1]=tim[x]+1; if(x-1==k)break;}
16         if(x<<1<=mx&&tim[x<<1]==-1){q.push(x<<1),tim[x<<1]=tim[x]+1; if(x<<1==k)break;}
17     }
18     printf("%d",tim[k]); return 0;
19 }

20160802

时间: 2024-10-09 23:41:28

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