Poj3278Catch That Cow (BFS)

Catch That CowTime Limit: 2000 MS Memory Limit: 65536 K Total Submissions: 37094 Accepted: 11466 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 teleport Ing. * 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 KOutput 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. there were a lot of strange YY at the beginning, so I don't know how many times it was rewritten .. Expand from the current location local to the local + 1, local-1, and local * 2 directions. Code:

#include <queue>   #include <cstdio>   #include<cstring>   #define N 100001   using namespace std;  int n, k, ans;  bool vis[N];  struct node {      int local, time;  };  void bfs() {      int t, i;      node now, tmp;      queue<node> q;      now.local = n;      now.time = 0;      q.push(now);      memset(vis,false,sizeof(vis));      vis[now.local] = true;      while(!q.empty()) {          now = q.front();          q.pop();          for(i=0; i<3; i++) {              if(0==i) t = now.local-1;              else if(1==i) t = now.local+1;              else if(2==i) t = now.local*2;              if(t<0||t>N||vis[t]) continue;              if(t==k) {                  ans = now.time+1;                  return;              }              vis[t] = true;              tmp.local = t;              tmp.time = now.time+1;              q.push(tmp);          }      }  }    int main() {      while(~scanf("%d%d",&n,&k)) {          if(n>=k) printf("%d\n",n-k);          else {              bfs();              printf("%d\n",ans);          }      }  }