HDU2717:Catch That Cow(BFS)
2013-06-03 21:54
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Problem 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
Sample Output
对所有状况进行一次搜索,最先找到的肯定就是时间最少的。
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 HintThe 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.
对所有状况进行一次搜索,最先找到的肯定就是时间最少的。
#include <stdio.h> #include <string.h> #include <queue> using namespace std; const int N = 1000000; int map[N+10]; int n,k; struct node { int x,step; }; int check(int x) { if(x<0 || x>=N || map[x]) return 0; return 1; } int bfs(int x) { int i; queue<node> Q; node a,next; a.x = x; a.step = 0; map[x] = 1; Q.push(a); while(!Q.empty()) { a = Q.front(); Q.pop(); if(a.x == k) return a.step; next = a; //每次都将三种状况加入队列之中 next.x = a.x+1; if(check(next.x)) { next.step = a.step+1; map[next.x] = 1; Q.push(next); } next.x = a.x-1; if(check(next.x)) { next.step = a.step+1; map[next.x] = 1; Q.push(next); } next.x = a.x*2; if(check(next.x)) { next.step = a.step+1; map[next.x] = 1; Q.push(next); } } return -1; } int main() { int ans; while(~scanf("%d%d",&n,&k)) { memset(map,0,sizeof(map)); ans = bfs(n); printf("%d\n",ans); } return 0; }
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