POJ 3278 Catch That Cow

题意:给一个起点和终点，只有三种走法n+1, n-1, n*2, 求最短需要的步数

解题思路:BFS，用BFS解决无权图的最短路问题。

注:很容易re，我re了10多次，有两个点要注意

1.如果起始点s>=终点e那么，就只能回退，所以step=s-e

2.编号不可能出边界(0, MAXSIZE)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#define MAXVERTEXNUM 1000100
using namespace std;

int dist[MAXVERTEXNUM];
int Visited[MAXVERTEXNUM];
int s, e;

void BFS()
{
    queue<int> q;
    q.push(s);
    Visited[s] = 1;

    while (!q.empty())
    {
        int num = q.front();
        if (num == e)
            return;
        q.pop();

        int d1 = num * 2, d2 = num + 1, d3 = num - 1;
        if (d1 < MAXVERTEXNUM && !Visited[d1])
        {
            dist[d1] = dist[num] + 1;
            q.push(d1);
            Visited[d1] = 1;
        }
        if (!Visited[d2])
        {
            dist[d2] = dist[num] + 1;
            q.push(d2);
            Visited[d2] = 1;
        }
        if (d3 >= 0 && !Visited[d3])
        {
            dist[d3] = dist[num] + 1;
            q.push(d3);
            Visited[d3] = 1;
        }
    }
}

int main()
{
    cin >> s >> e;
    memset(dist, 0, sizeof(dist));
    memset(Visited, 0, sizeof(Visited));

    if (s >= e)
        cout << s - e << endl;
    else
    {
        BFS();
        cout << dist[e] << endl;
    }

    return 0;
}