| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 222142 | Accepted: 67092 |
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 KOutput
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input
5 17
Sample Output
4
Hint
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. 题目大意： FJ抓位置固定不动的奶牛（二者均在一条直线上），FJ和奶牛位置分别为输入的N(0 ≤ N ≤ 100,000)和K(0 ≤ N ≤ 100,000)，FJ有两种走法*步行：FJ可以在一分钟内从任何一个点X移动到点X-1或X+1
*传送：FJ可以在一分钟内从任何一个X点移动到2*X点
输出FJ抓带到奶牛需要的最短时间，以分钟为单位
思路：
明显的BFS，几年没摸，重新回顾BFS，手写结构体构造先进先出的队列，
AC代码：
1 #include<stdio.h>
2 int iswalk[100050];
3 int p;//实时移动数组下标，视作队尾添加
4 int q;//取数，视作队头出队
5 int N, K;
6 struct trajectory
7 {
8 int step;
9 int num;
10 };
11 struct trajectory tra[100050];
12 int main()
13 {
14 scanf("%d %d", &N, &K);
15 tra[q].num = N;
16 tra[q].step = 0;
17 iswalk[N] = 1;
18 while(q <= p)
19 {
20 if(tra[q].num == K){
21 printf("%d\n", tra[q].step);
22 break;
23 }
24 else
25 {
26 if(iswalk[tra[q].num - 1] == 0 && (tra[q].num - 1 >= 0) && (tra[q].num - 1) <= 100000){
27 iswalk[tra[q].num - 1] = 1;
28 p ++;
29 tra[p].num = tra[q].num - 1;
30 tra[p].step = tra[q].step + 1;
31 }
32 if(iswalk[tra[q].num + 1] == 0 && (tra[q].num + 1 >= 0) && (tra[q].num + 1) <= 100000){
33 iswalk[tra[q].num + 1] = 1;
34 p ++;
35 tra[p].num = tra[q].num + 1;
36 tra[p].step = tra[q].step + 1;
37 }
38 if(iswalk[tra[q].num * 2] == 0 && (tra[q].num * 2) >= 0 && (tra[q].num * 2) <= 100000){
39 iswalk[tra[q].num * 2] = 1;
40 p ++;
41 tra[p].num = tra[q].num * 2;
42 tra[p].step = tra[q].step + 1;
43 }
44
45 q ++;
46 }
47 }
48 }
起手错误写法：
一：for的DFS（POJ3984的代码，都是BFS，于是推翻并找到奶牛这个基础题来写）
1 #include<stdio.h>
2 int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
3 int bfs(int x, int y);
4 int maze[5][5];
5 int step;
6 int evestep[25][2];//记录轨迹
7 int OnGo[5][5];//是否走过这个点
8 int main()
9 //int dir[4][2] = {(-1, 0), (1, 0), (0, -1), (0, 1)};
10 {
11 // int step;
12 for(int i = 0; i < 5; i ++)
13 for(int j = 0; j < 5; j ++)
14 scanf("%d", &maze[i][j]);
15 int x = 0, y = 0;
16 OnGo[0][0] == 1;
17 // int bfs(x, y);
18 bfs(x, y);
19 }
20 int bfs(int x, int y)
21 {
22 for(int i = 0; i < 4; i ++)
23 {
24 x += dir[i][0];
25 y += dir[i][1];
26 if(maze[x][y] == 1 || x < 0 || y < 0 || x > 4 || y > 4)
27 continue;
28 else if(maze[x][y] == 0 && OnGo[x][y] == 0)
29 {
30 OnGo[x][y] == 1;
31 step++;
32 evestep[step][0] = x;
33 evestep[step][1] = y;
34 bfs(x, y);
35 }
36 }
37 //}
38 //推翻重写，昨天写的是DFS
39
40 //BFS
View Code
二：推翻重写后，依旧是DFS的思路，以为BFS会携带每一层的step
1 #include<stdio.h>
2 int tra[100050];
3 int iswalk[100050];
4 int p, q;
5 int bfs(int FJ_x, int step)
6 {
7 if(FJ_x == K)
8 return step;
9 else
10 {
11 if((FJ_x - 1) >= 0 && iswalk[FJ_x - 1] == 0){
12 iswalk[FJ_x - 1] = 1;
13 step ++;
14 bfs(FJ_x - 1, step)
15 }
16 if((FJ_x + 1) >= 0 && iswalk[FJ_x + 1] == 0){
17 iswalk[FJ_x + 1] = 1;
18 step ++;
19 bfs(FJ_x + 1, step)
20 }
21 if((FJ_x * 2) >= 0 && iswalk[FJ_x * 2] == 0){
22 iswalk[FJ_x * 2] = 1;
23 step ++;
24 bfs(FJ_x * 2, step)
25 }
26 }
27 }
View Code
标签：Cow,int,POJ3278,bfs,tra,step,iswalk,Catch,FJ From： https://www.cnblogs.com/gerjcs/p/18307728