C - Catch That Cow POJ - 3278（一维BFS）

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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.

思路：基础BFS

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;const int maxn = 100000 + 5;
int dis[maxn];
int vis[maxn];
queue<int>Q;void bfs()
{while(!Q.empty()){int now = Q.front();Q.pop();for(int i = 0;i < 3;i++){int nex;if(i == 0)nex = now - 1;else if(i == 1)nex = now + 1;elsenex = now * 2;if(nex >=0 && nex <= 100000 && vis[nex] == 0){vis[nex] = 1;dis[nex] = 1 + dis[now];Q.push(nex);}}}
}int main()
{int n,m;while(~scanf("%d%d",&n,&m)){while(!Q.empty()){Q.pop();}if(n >= m){printf("%d\n",n - m);continue;}memset(vis,0,sizeof(vis));memset(dis,0,sizeof(dis));vis[n] = 1;Q.push(n);bfs();printf("%d\n",dis[m]);}return 0;
}

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