Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 82031 | Accepted: 25785 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
Source
用BFS即可,有地方需要注意,①visit数组要开到200000以上,否则“心灵传输”时可能会下标越界,②当上一个状态坐标超过K时不用向前走或“心灵传输”了,防止越界和优化时间。
# include <stdio.h> # include <queue> # include <string.h> using namespace std; struct node { int x, step; }; queue<node>Q; int vis[200001]; int bfs(int pos, int target) { node tmp; tmp.x = pos; tmp.step = 0; Q.push(tmp); while(!Q.empty()) { node tmp = Q.front(); if(tmp.x == target) return tmp.step; if(tmp.x > 0 && !vis[tmp.x-1]) { node temp; vis[tmp.x-1] = 1; temp.x = tmp.x-1; temp.step = tmp.step + 1; Q.push(temp); } if(tmp.x <= target && !vis[tmp.x+1]) { node temp; vis[tmp.x+1] = 1; temp.x = tmp.x+1; temp.step = tmp.step + 1; Q.push(temp); } if(tmp.x <= target && !vis[(tmp.x)<<1]) { node temp; vis[(tmp.x)<<1] = 1; temp.x = (tmp.x)<<1; temp.step = tmp.step + 1; Q.push(temp); } Q.pop(); } } int main() { int n, m; while(scanf("%d%d",&n,&m) != EOF) { if(m <= n) { printf("%d\n",n-m); continue; } memset(vis, 0, sizeof(vis)); while(!Q.empty()) Q.pop(); printf("%d\n",bfs(n, m)); } return 0; }