poj 3278 Catch That Cow

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?

Input

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

Output

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

Sample Input

5 17

Sample Output

4

Hint

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.

Source

USACO 2007 Open Silver

抽象的广度优先搜索。

队列q定义为全局变量时ac，但是放到bfs函数里面就会wa，可能队列中元素太多导致栈溢出，而全局变量不容易溢出。

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

const int size = 100001;
bool mark[size];
int dist[size];
queue<int> q;

int bfs(int N, int K) {
    q.push(N);
    dist[N] = 0;
    mark[N] = true;
    int i, v, w;
    while (!q.empty()) {
        v = q.front();
        q.pop();
        for (i = 0; i < 3; i++) {
            if (i == 0)
                w = v - 1;
            else if (i == 1)
                w = v + 1;
            else
                w = v * 2;
            if (w < 0 || w >= size)
                continue;
            if (!mark[w]) {
                mark[w] = true;
                q.push(w);
                dist[w] = dist[v] + 1;
            }
            if (w == K)
                return dist[w];
        }
    }
}

int main() {
    int N, K;
    scanf("%d%d", &N, &K);
    if (N >= K)
        printf("%d
", N - K);
    else
        printf("%d
", bfs(N, K));
    return 0;
}