POJ T3278 Catch That Cow 

题解：

    水题，Bfs裸题，变成一维更简单，走的方向只要判断下为2时是乘操作就行了......

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define maxn 100000   
using namespace std;

int n,k;
bool vis[maxn];
int mov[3] = {-1,1,2};  //三种走法

struct Node{
	int x,st;
}now,nex;

bool charge(int x){
	if(x < 0 || x > maxn || vis[x])   //边界判断
	    return false;
	return true;
}

int Bfs(int st){
	now.x = n;
	now.st = st;
	queue<Node> Q;
	Q.push(now);
	while(!Q.empty()){
		now = Q.front();
		Q.pop();
		if(now.x == k)
		    return now.st;
		for(int i = 0; i < 3; ++i){
			if(i == 2)
			    nex.x = now.x * mov[i];   //判断i == 2时为乘操作
			else
			    nex.x = now.x + mov[i];
			if(charge(nex.x)){
				nex.st = now.st + 1;
				vis[nex.x] = true;    //走过的点要标记掉
				Q.push(nex);
			}
		}
	}
}

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