POJ 3278 Catch That Cow

最新推荐文章于 2024-09-04 11:33:10 发布

lizhaowei213

最新推荐文章于 2024-09-04 11:33:10 发布

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分类专栏： BFS

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本文介绍了一个经典的算法问题：如何在最短的时间内通过加1、减1或乘2的操作将一个数字变为另一个数字。该问题使用了广度优先搜索(BFS)算法进行求解，并提供了一段完整的C++代码实现。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 87429		Accepted: 27431

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

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

从一个数n到另一个数k，有三种操作，加1减1乘2，问最少几步能把n变成k

BFS水题，温暖一下。

#include <cstdio>
#include <cstring>
#include <queue>
#define MAXN 100000+5
using namespace std;

int step[MAXN];
int visited[MAXN];
queue<int> nodeQueue;

int bfs(int n,int k){
    while(!nodeQueue.empty())
        nodeQueue.pop();
    step[n]=0;
    visited[n]=1;
    nodeQueue.push(n);
    while(!nodeQueue.empty()){
        int current=nodeQueue.front();
        nodeQueue.pop();
        int next;
        for(int i=0;i<3;i++){
            if(i==0)
                next=current-1;
            if(i==1)
                next=current+1;
            if(i==2)
                next=current*2;
            if(next<0||next>MAXN)
                continue;
            if(!visited[next]){
                nodeQueue.push(next);
                step[next]=step[current]+1;
                visited[next]=1;
            }
            if(current==k)
                return step[current];
        }
    }
}

int main() {
    int n,k;
    while(scanf("%d%d",&n,&k)>0){
        memset(step,0, sizeof(step));
        memset(visited,0, sizeof(visited));
        if(n>=k)
            printf("%d\n",n-k);
        else
            printf("%d\n",bfs(n,k));
    }
    return 0;
}