标签：des style blog io ar color sp for strong
| Time Limit: 2000MS | Memory Limit: 65536K | |
Description
Farmer John has been informed of the location of a fugitive（逃亡的;难以捉摸的;短暂的) cow and wants to catch her immediately. He starts at a pointN(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 orX+ 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(追赶;工作), 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
题意：在一维坐标系中，给定John的位置，还有cow的位置，john有2种行动方式，（一）向前走一步（-1），想后走一步（+1）；（二）跳到当前位置的2倍位置。
问：最少需要多少时间(分钟)可以抓到cow。
【用vis数组优化】：如果不加“代码中阴影”，会RE
【bfs搜索即可】：
#include<stdio.h> #include<string.h> int dx[3]={-1, 1, 2}; int vis[3000000]; int n, k; struct node{ int x, ans; }q[3000000], t, f; void bfs() { memset(vis, 0, sizeof(vis)); memset(q, 0, sizeof(q)); int s=0, e=0; vis[n] = 1; q[s++].x = n; t.ans = 0; while(s>e) { t = q[e++]; for(int i=0; i<3; i++) { if(i!=2) f.x = t.x+dx[i]; else f.x = t.x*dx[i]; f.ans = t.ans+1; if(f.x==k) { printf("%d\n", f.ans); return; } else{ if(!vis[f.x] && (f.x>0 && f.x<100000) ) { vis[f.x] = 1; q[s++] = f; } } } } } int main() { while(~scanf("%d%d", &n, &k)) { if(n>=k) { printf("%d\n", n-k); continue; } else { bfs(); } } return 0; }
标签：des style blog io ar color sp for strong
原文地址：http://www.cnblogs.com/6bing/p/4118948.html
