POJ 3278 Catch That Cow

解题思路：简单宽搜，关键是剪枝。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 100005;
int vis[maxn]; //标记这个点是否走过
int n, k;


struct node{
    int x, cnt;
}p, now;

queue<node> q;

int bfs(int x, int cnt)
{
    memset(vis, 0, sizeof(vis)); //初始化为没走过
    while(!q.empty()) q.pop();
    p.x = x, p.cnt = cnt;
    q.push(p);

    while(!q.empty())
    {
        p = q.front();
        q.pop();

        if(p.x == k) return p.cnt; //走到该点，直接返回步数

        now.x = p.x + 1, now.cnt = p.cnt + 1;
        if(now.x <= 100000 && !vis[now.x]) //关键剪枝，没走过并且在符合题目的数据范围之内
        {
            vis[now.x] = 1; //标记为已经走过
            q.push(now);
        }

        now.x = p.x - 1, now.cnt = p.cnt + 1;
        if(now.x >= 0 && !vis[now.x])
        {
            vis[now.x] = 1;
            q.push(now);
        }

        now.x = 2 * p.x, now.cnt = p.cnt + 1;
        if(now.x <= 100000 && !vis[now.x])
        {
            vis[now.x] = 1;
            q.push(now);
        }
    }
    return -1; //不会走到这一步
}

int main()
{
    while(~scanf("%d %d", &n, &k))
    {
        int ans = bfs(n, 0);
        printf("%d
", ans);
    }
    return 0;
}