kuangbin一C：HDU2717 Catch That Cow（BFS）

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

5 17

Sample Output

4

这个题与B相比比较简单，BFS。

#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

const int N = 1000000;
int vis[N+10];
int n,k;
int to[3] = {1, -1, 0};

struct loc
{
    int x, level;
};

int checktrue(int num)
{
    if(num<0 || num>=N || vis[num])
        return 0;
    return 1;
}

int bfs(int m)
{
    queue<loc> srch;
    loc curNum, nxt;
    curNum.x = m;
    curNum.level = 0;
    vis[m] = 1;
    srch.push(curNum);
    while(!srch.empty())
    {
        curNum = srch.front();
        srch.pop();
        if(curNum.x == k)
            return curNum.level;
        nxt = curNum;
        to[2] = curNum.x;
        for(int i = 0; i < 3; i++)
        {
            nxt.x = curNum.x + to[i];
            if(checktrue(nxt.x))
            {
                nxt.level = curNum.level+1;
                vis[nxt.x] = 1;
                srch.push(nxt);
            }
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%d %d", &n, &k))
    {
        memset(vis, 0, sizeof(vis));
        printf("%d\n", bfs(n));
    }
    return 0;
}