POJ 3278 广度搜索 一个终点

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 66244 Accepted: 20811

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.

遇到的问题和思路：

       刚开始看这道题目的时候，第一反应就是深度搜索，然后仔细看了一下并没有发觉如何搜索，因为深度搜索是分顺序的，很难描述（或许有些大神会，这里不要吐槽我哦）。然后就想到了广度搜索，用队列。然后到广搜的时候又碰到了问题，因为看了一下别人的思路和我一样，然而我TLE ，别人是AC，就是因为别人在push到队列里去的时候if中多了一个条件，然而我很不解为什么要多这一个条件，后来岂不是可能也走到这个位置吗。然后想了一想，确实，如果之前能走到这个位置呢，那么用的时间肯定比后来访问到的要短，那么还要后来访问到的干什么呢。（这就是一种优化）

给出代码：

#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<queue>#define inf 1000000000using namespace std;int n, k;int map[100005];//typedef pair<int, int> P;queue <int> que;int bfs(){que.push(n);while(que.size()){int x = que.front();que.pop();if(x == k)break;if(x > k){que.push(x - 1);map[x - 1] = min(map[x - 1], map[x] + 1);continue;}if(x > 0 && x <= 100000 && map[x - 1] == inf){que.push(x - 1);map[x - 1] = min(map[x] + 1, map[x - 1]);}if(x >= 0 && x < 100000 && map[x + 1] == inf){que.push(x + 1);map[x + 1] = min(map[x + 1], map[x] + 1);}if(x <= 50000 && x > 0 && map[x * 2] == inf){que.push(x * 2);map[x * 2] = min(map[x * 2], map[x] + 1);}}//printf("%d\n", map[k]);return 0;}void solve(){    bfs();printf("%d\n", map[k]);}int main(){while(scanf("%d%d",&n,&k)!=EOF){for(int i = 0;i <= 100003; i++)map[i] = inf;map[n] = 0;solve();}return 0;}