【搜索算法】练习二：poj3278 Catch That Cow、poj1426 Find The Multiple

poj3278 Catch That Cow

题目链接：https://vjudge.net/problem/POJ-3278

1.题目描述

  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?

2.输入要求

  Line 1: Two space-separated integers: N and K

3.输出要求

  Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

4.题目解释

  农夫约翰已被告知一头逃犯的位置，并希望立即将其抓住。他开始于一个点Ñ（0≤ Ñ在数轴上≤100,000）和母牛是在点ķ（0≤ ķ上相同数目的线≤100,000）。农夫约翰有两种运输方式：步行和传送。

  行走：FJ可以在一分钟内从任意点X移至点X -1或X + 1。
  传送：FJ可以在一分钟内从任意点X移至点2× X。

  如果没有意识到它的追捕能力的母牛完全没有动弹，那么农夫约翰要花多长时间才能找回它？

5.测试样例

Sample Input
5 17

Sample Output
4

6.代码

#include
#include
#include
using namespace std;

//n,k最大为100000
#define N 200001
bool visit[N];
int n,k;

//结构体，点和步数
struct node{
    int id;
    int step;
};

//广度优先遍历
int bfs(int n,int k)
{
    queue<node> q;
    node f;
    //初始化
    f.id = n;
    f.step = 0;
    visit[n] = true;
    q.push(f);
    while(!q.empty())
    {
        node now = q.front();
        q.pop();
        if(now.id==k) return now.step;
        //分别遍历三种情况
        node next;
        for(int i=0;i<3;i++)
        {
            if(i==0) next.id = now.id - 1;
            else if(i==1) next.id = now.id + 1;
            else next.id = now.id * 2;

            if(next.id<0 || next.id>=N || visit[next.id]) continue;
            else
            {
                next.step = now.step + 1;
                visit[next.id] = true;
                q.push(next);
            }
        }
    }
}

int main()
{
    cin>>n>>k;
    memset(visit,false,sizeof(visit));
    cout<<bfs(n,k);
    return 0;
}

---

poj1426 Find The Multiple

题目链接：https://vjudge.net/problem/POJ-1426

1.题目描述

  Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

2.输入要求

  The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

3.输出要求

  For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

4.题目解释

  给定正整数n，编写一个程序以找出n的非零倍数m，其十进制表示形式仅包含数字0和1。您可以假定n不大于200，并且对应的m包含不超过100十进制数字。

5.测试样例

Sample Input
2
6
19
0

Sample Output
10
100100100100100100
111111111111111111

6.代码

#include
using namespace std;

int n,flag;

//广度优先遍历
void bfs(long long num,int sum)
{
    //long long最长为19位
    if(flag || sum==19)
    {
        return;
    }
    if(num%n==0)
    {
        cout<<num<<endl;
        flag = 1;
        return;
    }
    //最后一位扩展为0或者1
    bfs(num*10,sum+1);
    bfs(num*10+1,sum+1);
}

int main()
{
    while(cin>>n)
    {
        if(n==0) break;
        else
        {
            flag = 0;
            bfs(1,0);
        }
    }
    return 0;
}