POJ 3278 Catch that Cow (BFS)

Transmission Door Catch that Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 80273		Accepted: 25290

Description

Farmer John had 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 are at a point K (0≤ K ≤100,000) on the same number line. Farmer John has 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 2x 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?

InputLine 1:two space-separated integers: N and K OutputLine 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

Hintthe fastest-Farmer John to reach the fugitive cow are to move along the following path:5-10-9-18-17, which Tak Es 4 minutes.Ideas

Test Instructions: given the starting and ending points, there are three ways to go, assuming that now, then you can choose the next time to reach Now-1 or now + 1 or 2*now, ask the beginning to the end of the minimum number of steps required.

Solving :BFS, the next state is three selectable positions.

#include <iostream> #include <cstdio> #include <cstring>using namespace std;const int maxn = 100005;int Step[maxn],que[maxn],vis[maxn];int BFS (int st,int ed) {int E = 0,f = 0;que[f++] = st;vis[st] = 1;for (;;) {int now = que[e];if (today = ed) return Step[now];if (now + 1 < MAXN &&!vis[now + 1]) Step[now + 1] = Step[now] + 1,vis[now + 1] = 1,que[f++] = now + 1;if (now-1 >= 0 &&!vis[now-1]) step[now-1] = Step[now] + 1,vis[now  -1] = 1,que[f++] = now-1;if (2*now < MAXN &&!vis[2*now]) Step[2*now] = Step[now] + 1,vis[2*now] = 1,que[f++] = 2 * NOW; e++;}} int main () {int N,k;memset (vis,0,sizeof (Vis)), scanf ("%d%d", &n,&k), if (N >= K) printf ("%d\n", n-k); elseprintf ("%d\n", BFS (N,k)); return 0;}

　　

POJ 3278 Catch that Cow (BFS)