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BFS PKU 3278
一道很简单的,典型的BFS。
代码虐我千百遍,我待代码如初恋
Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 47366 | Accepted: 14869 |
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 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.
Source
USACO 2007 Open Silver
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <ctype.h> #include <limits.h> #include <string.h> #include <string> #include <math.h> #include <algorithm> #include <iostream> #include <queue> #include <stack> #include <vector> #include <set> #include <map> #include <deque> using namespace std; #define MAXN 100000 + 10 int vis[MAXN]; int k; struct node{ int num; int sum; //int vis; }a[MAXN]; void BFS(int n){ queue<node>q; node front; node rear; front.num = n; vis[n] = 1; front.sum = 0; q.push(front); int mark1; int mark2; int mark3; //int mark4; //int sum = 0; while(!q.empty()){ front = q.front(); q.pop(); if(front.num == k){ printf("%d\n",front.sum); break; } mark1 = front.num - 1; mark2 = front.num + 1; mark3 = front.num*2; if(mark1>=0 && vis[mark1]==0){ rear.num = mark1; rear.sum = front.sum+1; //sum++; vis[mark1] = 1; q.push(rear); } if(mark2<=100000 && vis[mark2]==0){ rear.num = mark2; //sum++; rear.sum = front.sum + 1; vis[mark2] = 1; q.push(rear); } if(mark3<=100000 && vis[mark3]==0){ rear.num = mark3; //sum++; rear.sum = front.sum + 1; vis[mark3] = 1; q.push(rear); } } } int main(){ int n; while(~scanf("%d%d",&n,&k)){ memset(vis,0,sizeof(vis)); //memset(a,0,sizeof(a)); BFS(n); } return 0; }
BFS PKU 3278
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