Catch That Cow Time Limit: 2000MS Memory Limit: 65536K 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 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 17Sample Output 4Hint 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. 剛做的時候發生了很多奇異YY, 搞的我不知道重寫了多少遍。。 從當前位置local,向local+1, local-1, local*2這三個方向擴展。 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); } } }