POJ 3278 Catch That Cow （BFS）

题目链接：Catch That Cow

解析：两个数n和k，三种操作：+1、-1、*2，问n最少经过多少次操作能和k相等。

最简单的bfs模板了，注意

+1的条件：x+1 <= k

-1的条件：x-1 >= 0

*2的条件：x <= k && 2*x <= 100000

AC代码：

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

struct Node { int n, step; };

bool vis[100005];
int bfs(int n, int k){
    memset(vis, false, sizeof(vis));
    queue<Node> Q;
    Q.push(Node{n, 0});
    while(!Q.empty()){
        Node now = Q.front();
        Q.pop();
        if(now.n == k) return now.step;
        if(now.n+1 <= k && !vis[now.n+1]){
            vis[now.n+1] = true;
            Q.push(Node{now.n+1, now.step+1});
        }
        if(now.n-1 >= 0 && !vis[now.n-1]){
            vis[now.n-1] = true;
            Q.push(Node{now.n-1, now.step+1});
        }
        if(now.n <= k && 2*now.n <= 100000 && !vis[2*now.n]){
            vis[2*now.n] = true;
            Q.push(Node{2*now.n, now.step+1});
        }
    }
    return -1;
}

int main(){
//    freopen("in.txt", "r", stdin);
    int n, k;
    while(scanf("%d%d", &n, &k) == 2){
        printf("%d\n", bfs(n, k));
    }
    return 0;
}

POJ 3278 Catch That Cow （BFS）

原文：http://blog.csdn.net/u013446688/article/details/47778391