poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 47033		Accepted: 14770

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?

John农民终于知道那个逃跑的母牛去哪了，并且想赶紧给抓回来。他在点N(0 ≤ N ≤ 100,000)，牛在点K (0 ≤ K ≤ 100,000) ，他们都在同一个数轴上。John有两种移动方式：走路和远距离传送。

走路：FJ可以从X移动到X-1和X+1在一分钟内

远距离传送：FJ可以从X移动到2*X在一分钟内

牛，还没意识到自己被跟踪了，根本不动，John多久才能逮到牛？

Input

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

两个数，N和K

Output

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

需要的时间

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

虽然用了BFS不过果然这道题除了二倍就是左右走，最好的办法，还是找到其中的规律。。。这个在discuss里面有，就是对N-K的逆推，效率十分高。。。

下面是BFS版本，需要对边界有点注意，坑了好几次。。。

#include <stdio.h>
#include <string.h>


struct Point{
    int dis;
    int step;
};
int main(void){

    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int head = 0, tail = 1;
    int visit[200004];
    int n, k;
    struct Point fifo[200004];

    scanf("%d%d", &n, &k);
    memset(visit, 0, sizeof(visit));

    visit[n] = 1;
    //printf("%d%d\n", n, k);

    fifo[head].dis = n;
    fifo[head].step = 0;

    while(head <= tail){
        struct Point curr = fifo[head];
        if (curr.dis == k){
            printf("%d\n", fifo[head].step);
            break;
        }

        if ((0 <= curr.dis - 1)  && (visit[curr.dis - 1] == 0)){
            fifo[tail].dis = curr.dis - 1;
            fifo[tail++].step = curr.step + 1;
            visit[curr.dis - 1] = 1;
            printf("tail = %d\n", tail);
        }

        if ((curr.dis <= k)  && (visit[curr.dis + 1] == 0)){//here not corr.dis + 1 <= k;
            fifo[tail].dis = curr.dis + 1;
            fifo[tail++].step = curr.step + 1;
            visit[curr.dis + 1] = 1;
        }

        if ((curr.dis <= k) && (visit[curr.dis * 2] == 0)){//here not corr.dis * 2 <= k;
            fifo[tail].dis = curr.dis * 2;
            fifo[tail++].step = curr.step + 1;
            visit[curr.dis * 2] = 1;
        }
        head ++;
    }
    return 0;
}