POJ 3278 Catch That Cow（赶牛行动）
Time Limit: 1000MS Memory Limit: 65536K
Description - 题目描述
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John被告知有奶牛企图逃跑欲火速擒回。他的初始位置为某条线的N (0 ≤ N ≤ 100,000)点，且奶牛在同一条线上的点K (0 ≤ K ≤ 100,000)。农夫John有两种移动方式：要么走路，要么传送。 * 走路: FJ可以从任意的点X用一分钟移动到 X - 1 或 X + 1 * 传送: FJ可以从任意的点X用一分钟传送到2 × X 如果牛完全不动，农夫John需要多少时间才能追回它们？
Input - 输入
Line 1: Two space-separated integers: N and K
1行：两个由空格分隔的整数：N和K
Output - 输出
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
1行：最短时间，表示农夫John需要多少分钟追回逃跑的奶牛。
Sample Input - 输入样例
5 17
Sample Output - 输出样例
4
题解
判断好条件，开辟两倍的空间（*2），直接SPFA（BFS）。
代码 C++
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 int data[200005]; 5 int main(){ 6 int n, k, now, nxt, i, j; 7 scanf("%d%d", &n, &k); 8 k <<= 1; 9 memset(data, 0x7F, sizeof data); data[n] = 0; 10 std::queue<int>q; q.push(n); 11 while (!q.empty()){ 12 now = q.front(); q.pop(); 13 j = now == 0 ? 1 : 3; 14 for (i = 0; i < j; ++i){ 15 switch (i){ 16 case 0:nxt = now + 1; break; 17 case 1:nxt = now << 1; break; 18 case 2:nxt = now - 1; break; 19 } 20 if (i<2 && nxt > k) continue; 21 if (data[nxt] > data[now] + 1){ 22 data[nxt] = data[now] + 1; q.push(nxt); 23 } 24 } 25 } 26 printf("%d ", data[k >> 1]); 27 return 0; 28 }