POJ3278解题报告

import java.io.BufferedInputStream;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

/*POJ3278
* 大致题意：
给定两个整数n和k
通过 n+1或n-1 或n*2 这3种操作，使得n==k
输出最少的操作次数
*
* */
public class Main {
static int MAX = 200000;

public static void main(String args[]) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int n = sc.nextInt();
int k = sc.nextInt();
System.out.println(bfs(n, k));
}

private static int bfs(int n, int k) {
// TODO Auto-generated method stub
Queue<Integer> queue = new LinkedList<Integer>();
queue.offer(n);
if (n == k) {
return 0;
}
int[] min = new int[MAX + 5];
boolean[] visited = new boolean[MAX + 5];
visited
= true;
while (!queue.isEmpty()) {
int next = queue.remove();
for (int i = 0; i < 3; i++) {
int temp = next;
if (i == 0) {
temp++;
} else if (i == 1) {
temp--;
} else if (i == 2) {
temp *= 2;
}
if (!visited[temp]) {
queue.offer(temp);
visited[temp] = true;
min[temp] = min[next] + 1;
}
if (temp == k) {
return min[k];
}
}
}
return 0;
}
}