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Catch That Cow BFS求线性双向最短路径
Catch That Cow
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
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.
开始考虑用dfs做。。没考虑双向问题，向左向右会陷入死循环。还是BFS大法好，，求最短路径问题，遍历部分点即可。
#include<stdio.h> #include<queue> using namespace std; struct Node{ int x,y; }node; int b[200010]; int main() { int n,k; queue<Node> q; scanf("%d%d",&n,&k); if(n==k) printf("0\n"); //特判 else{ node.x=n; node.y=0; q.push(node); b[n]=1; while(q.size()){ node.x=q.front().x*2; node.y=q.front().y+1; if(node.x<=200010&&node.x>=0){ if(node.x==k){ printf("%d\n",node.y); break; } if(b[node.x]==0){ b[node.x]=1; q.push(node); } } node.x=q.front().x+1; node.y=q.front().y+1; if(node.x<=200010&&node.x>=0){ if(node.x==k){ printf("%d\n",node.y); break; } if(b[node.x]==0){ b[node.x]=1; q.push(node); } } node.x=q.front().x-1; node.y=q.front().y+1; if(node.x<=200010&&node.x>=0){ if(node.x==k){ printf("%d\n",node.y); break; } if(b[node.x]==0){ b[node.x]=1; q.push(node); } } q.pop(); } } return 0; }
Catch That Cow BFS求线性双向最短路径
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