[BZOJ1646][Usaco2007 Open]Catch That Cow 抓住那只牛

1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 1271  Solved: 601 [Submit][Status][Discuss]

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

4

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.

 

bfs

#include <queue>
#include <cstdio>
using namespace std;
int n, k, f[2000000 + 10];
queue<int> q;
bool inq[2000000 + 10] = {false};
void bfs(){
    f[n] = 0;
    q.push(n);
    inq[n] = true;
    int u, v;
    while(!q.empty()){
        u = q.front(); q.pop();
        if(u == k) return;
        if(u > k){
            v = u - 1;
            if(!inq[v]){
                inq[v] = true;
                f[v] = f[u] + 1;
                q.push(v);
            }
        }
        else{
            v = u - 1;
            if(v > 0 && !inq[v]){
                inq[v] = true;
                f[v] = f[u] + 1;
                q.push(v);
            }
            v = u + 1;
            if(!inq[v]){
            inq[v] = true;
            f[v] = f[u] + 1;
            q.push(v);
            }
            v = u * 2;
            if(!inq[v]){
                inq[v] = true;
                f[v] = f[u] + 1;
                q.push(v);
            }
        }
    }
}
int main(){
    scanf("%d %d", &n, &k);
    bfs();
    printf("%d\n", f[k]);
    return 0;
}