POJ 3278:Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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
发现广度优先搜索适合什么题呢?就是那种在每个点都给你几个选择,然后问你最短路径的问题,对,就是这样,这样的题目最适合广度优先搜索。
这次的这个就是,每次Farmer有三个选择,然后求最短到达目的地的行程。简直是广搜的模板题。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <queue> #include <cstring> #pragma warning(disable:4996) using namespace std; int color[1000005]; int dis[1000005]; queue<int> q; int main() { //freopen("i.txt","r",stdin); //freopen("o.txt","w",stdout); int N,K; cin>>N>>K; if(N==K) { cout<<0<<endl; } else { memset(color,0,sizeof(color)); memset(dis,0,sizeof(dis)); q.push(N); while(!q.empty()) { N=q.front(); q.pop(); if(N-1==K || N+1==K || 2*N==K) { cout<<dis[N]+1<<endl; break; } if(color[N-1]==0 && N-1>=0) { color[N-1]=1; dis[N-1]=dis[N]+1; q.push(N-1); } if(color[N+1]==0) { color[N+1]=1; dis[N+1]=dis[N]+1; q.push(N+1); } if(color[2*N]==0 && 2*N<=100000) { color[2*N]=1; dis[2*N]=dis[N]+1; q.push(2*N); } } } return 0; }
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