POJ 3278： Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 44613		Accepted: 13946

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

复制代码5 17

1
5 17

Sample Output

复制代码4

1
4

题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户開始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1。2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次能够移动到牛所在的位置。所以能够用BFS来搜索这三个状态，直到搜索到牛所在的位置。

复制代码#include<cstdio> #include<iostream> #include<cstring> #include<queue> #include<algorithm> using namespace std; const int N = 200100; int n, k; struct node { int x, step; }; queue<node> q; int vist[N]; void bfs() { int cow, ans; while(!q.empty()) { node tmp = q.front(); q.pop(); cow = tmp.x; ans = tmp.step; if(cow == k) { printf("%dn",ans); return ; } if(cow >= 1 && !vist[cow - 1]) //要保证减1后有意义,所以要cow >= 1 减一的情况 { node temp; vist[cow - 1] = 1; temp.x = cow - 1; temp.step = ans + 1; q.push(temp); } if(cow <= k && !vist[cow + 1]) //加1的情况 { node temp; vist[cow + 1] = 1; temp.x = cow + 1; temp.step = ans + 1; q.push(temp); } if(cow <= k && !vist[cow * 2]) //乘二的情况 { node temp; vist[cow * 2] = 1; temp.x = 2 * cow; temp.step = ans + 1; q.push(temp); } } } int main() { while(~scanf("%d%d",&n,&k)) { while(!q.empty()) q.pop(); memset(vist,0,sizeof(vist)); vist[n] = 1; node t; t.x = n, t.step = 0; q.push(t); bfs(); } return 0; }

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 200100;
int n, k;
struct node
{
int x, step;
};
queue<node> q;
int vist[N];
void bfs()
{
int cow, ans;
while(!q.empty())
{
node tmp = q.front();
q.pop();
cow = tmp.x;
ans = tmp.step;
if(cow == k)
{
printf("%dn",ans);
return ;
}
if(cow >= 1 && !vist[cow - 1]) //要保证减1后有意义,所以要cow >= 1
减一的情况
{
node temp;
vist[cow - 1] = 1;
temp.x = cow - 1;
temp.step = ans + 1;
q.push(temp);
}
if(cow <= k && !vist[cow + 1]) //加1的情况
{
node temp;
vist[cow + 1] = 1;
temp.x = cow + 1;
temp.step = ans + 1;
q.push(temp);
}
if(cow <= k && !vist[cow * 2]) //乘二的情况
{
node temp;
vist[cow * 2] = 1;
temp.x = 2 * cow;
temp.step = ans + 1;
q.push(temp);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
while(!q.empty()) q.pop();
memset(vist,0,sizeof(vist));
vist[n] = 1;
node t;
t.x = n, t.step = 0;
q.push(t);
bfs();
}
return 0;
}