BZOJ 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛 (BFS)
1646: [Usaco2007 Open]Catch That Cow 抓住那只牛
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 915 Solved: 441 [Submit][Status][Discuss]
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?
农夫约翰被通知,他的一只奶牛逃逸了!所以他决定,马上幽发,尽快把那只奶牛抓回来.
他们都站在数轴上.约翰在N(O≤N≤100000)处,奶牛在K(O≤K≤100000)处.约翰有
两种办法移动,步行和瞬移:步行每秒种可以让约翰从z处走到x+l或x-l处;而瞬移则可让他在1秒内从x处消失,在2x处出现.然而那只逃逸的奶牛,悲剧地没有发现自己的处境多么糟糕,正站在那儿一动不动.
那么,约翰需要多少时间抓住那只牛呢?
Input
- Line 1: Two space-separated integers: N and K
仅有两个整数N和K.
Output
- Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
最短的时间.
Sample Input
5 17 Farmer John starts at point 5 and the fugitive cow is at point 17.
Sample Output
4
OUTPUT DETAILS:
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.
大概是写得第一道bfs了。
/* ***********************************************
Author :111qqz
Created Time :2016年04月10日 星期日 21时00分01秒
File Name :code/bzoj/1646.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E5+7;
7int n,k;
8bool vis[N];
9int d[N];
10void bfs()
11{
12 ms(d,-1);
13 queue<int>q;
14 q.push(n);
15 vis[n] = true;
16 d[n] = 0 ;
1 while (!q.empty())
2 {
3 int px = q.front () ; q.pop();
4 if (px==k)
5 {
6 return ;
7 }
8 int nxt[3];
9 nxt[0] = px-1;
10 nxt[1] = px+1;
11 nxt[2] = 2*px;
12 for ( int i = 0 ; i < 3 ; i++)
13 {
14 if (nxt[i]>=0&&nxt[i]<=100000&&!vis[nxt[i]])
15 {
16 q.push(nxt[i]);
17 vis[nxt[i]] = true;
18 d[nxt[i]] = d[px] + 1;
19 }
20 }
21 }
22}
23int main()
24{
25 #ifndef ONLINE_JUDGE
26 freopen("code/in.txt","r",stdin);
27 #endif
1 scanf("%d %d",&n,&k);
2 bfs();
3 printf("%d\n",d[k]);
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}