标签：spl iat play sep gif 技术分享 dia int win
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?
Line 1: Two space-separated integers: N and K
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
5 17
4
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.
以前的都是二维，突然一维有点不会qaq
不能超范围，n==k时要特殊判断。
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 using namespace std; 8 int n, k; 9 int book[100010]; 10 struct student 11 { 12 int x, step; 13 }que[100010]; 14 bool check(int x) 15 { 16 if (x < 0 || x>100000 || book[x] == 1) 17 return 0; 18 return 1; 19 } 20 int bfs() 21 { 22 if (n == k) return 0; 23 book[n] = 1; 24 int head = 1, tail = 2; 25 que[head].x = n; 26 que[head].step = 0; 27 while (head < tail) { 28 int tx1 = que[head].x - 1; 29 if (check(tx1)) 30 { 31 que[tail].x = tx1; 32 que[tail].step = que[head].step + 1; 33 book[tx1] = 1; 34 if (tx1 == k) 35 return que[tail].step; 36 tail++; 37 } 38 int tx2 = que[head].x + 1; 39 if (check(tx2)) 40 { 41 que[tail].x = tx2; 42 que[tail].step = que[head].step + 1; 43 book[tx2] = 1; 44 if (tx2 == k) 45 return que[tail].step; 46 tail++; 47 } 48 int tx3 = que[head].x * 2; 49 if (check(tx3)) 50 { 51 que[tail].x = tx3; 52 que[tail].step = que[head].step + 1; 53 book[tx3] = 1; 54 if (tx3 == k) 55 return que[tail].step; 56 tail++; 57 } 58 head++; 59 } 60 } 61 int main() 62 { 63 cin >> n >> k; 64 cout << bfs() << endl; 65 return 0; 66 }
标签：spl iat play sep gif 技术分享 dia int win
原文地址：https://www.cnblogs.com/Kaike/p/9914385.html
