BZOJ1646[Usaco2007 Open] 抓住那头牛

原题链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1646
一本通链接：http://ybt.ssoier.cn:8088/problem_show.php?pid=1253

抓住那头牛

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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

那么，约翰需要多少时间抓住那只牛呢？

Input

Line 1: Two space-separated integers: N and K

仅有两个整数N和K.

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

最短的时间．

Sample Input

5 17

Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

OUTPUT DETAILS:

The fastest way for Farmer John to reach the fugitive cow is to

move along the following path: 5-10-9-18-17, which takes 4 minutes.

题解

bfs裸题，巨水不解释。

代码

#include<bits/stdc++.h>
using namespace std;
const int M=1e5+5;
int p1=1,p2=1;
int n,k;
bool sq[M];
int step[M][2];
int main()
{scanf("%d%d",&n,&k);sq[n]=1;step[1][1]=n;int maxn=max(n,k)+1;if(n==k)printf("0"),exit(0);while(p1<=p2){int a;for(int i=1;i<=3;i++){if(i==1)a=step[p1][1]-1;if(i==2)a=step[p1][1]+1;if(i==3)a=step[p1][1]*2;if(a<0||a>maxn)continue;if(sq[a]==0){p2++;sq[a]=1;step[p2][1]=a;step[p2][0]=step[p1][0]+1;if(a==k){printf("%d\n",step[p2][0]);return 0;}}}p1++;}return 0;
}

凯老师神奇的dp版

#include<cstdio>
#include<queue>
using namespace std;
int dp[100005];
int n,k;
struct no{int pl,step;
};
queue<no> q;
void BFS(){q.push((no){n,1});dp[n]=1;while(!q.empty()){if(dp[k]!=0) {printf("%d",dp[k]-1);return;}no u=q.front();q.pop();if(u.pl+1<=100000&&dp[u.pl+1]==0){dp[u.pl+1]=u.step+1;q.push((no){u.pl+1,u.step+1});}if(u.pl-1>=0&&dp[u.pl-1]==0){dp[u.pl-1]=u.step+1;q.push((no){u.pl-1,u.step+1});}if((u.pl<<1)<=100000&&dp[u.pl<<1]==0){dp[u.pl<<1]=u.step+1;q.push((no){u.pl<<1,u.step+1});}}
}
int main(){scanf("%d%d",&n,&k);BFS();return 0;
}