POJ 3278 Catch That Cow(模板——BFS)
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题目链接:
http://poj.org/problem?id=3278
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
Output
Sample Input
5 17
Sample Output
4
Hint
输入人和牛在坐标轴上的位置
1 #include<stdio.h> 2 int bfs(int n,int k); 3 struct node 4 { 5 int x,s; 6 }; 7 struct node q[300010]; 8 int book[100001];//标记数组,否则超内存 9 int main() 10 { 11 int n,k; 12 while(scanf("%d%d",&n,&k) != EOF) 13 { 14 if(n==k) 15 printf("0\n"); 16 else 17 printf("%d\n",bfs(n,k)); 18 } 19 return 0; 20 } 21 int bfs(int n,int k) 22 { 23 int i,head,tail,tx; 24 head=1; 25 tail=1; 26 q[tail].x=n; 27 q[tail].s=0; 28 book[n]=1; 29 tail++; 30 while(head < tail) 31 { 32 for(i=1;i<=3;i++) 33 { 34 if(i==1) 35 tx=q[head].x-1; 36 if(i==2) 37 tx=q[head].x+1; 38 if(i==3) 39 tx=q[head].x*2; 40 if(tx < 0||tx > 100000) 41 continue; 42 if(!book[tx]) 43 { 44 book[tx]=1; 45 q[tail].x=tx; 46 q[tail].s=q[head].s+1; 47 tail++; 48 } 49 if(tx == k) 50 return q[tail-1].s; 51 } 52 head++; 53 } 54 }
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