C - 开
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2717
Description
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?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
Hint
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. 之所以把这个水题贴上,是为了警戒自己,数组超限与范围判断的时候,应该先进行范围判断,然后进行数组操作!
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; int vis[300010]; int bfs(int x,int ans){ queue<int>q; queue<int>num; q.push(x); num.push(0); while(!q.empty()){ int xx = q.front(); int nn = num.front(); num.pop(); q.pop(); if(xx == ans)return nn; if(xx>=0&&vis[xx-1] == 0){ q.push(xx-1); num.push(nn+1); vis[xx-1] = 1; } if(xx+1<=ans&&vis[xx+1] == 0){ q.push(xx+1); num.push(nn+1); vis[xx+1] = 1; } if(2*xx <=ans*2&&vis[2*xx] == 0){ q.push(2*xx); num.push(nn+1); vis[2*xx] = 1; } } } int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ memset(vis,0,sizeof(vis)); vis[n]=1; printf("%d\n",bfs(n,m)); } }