POJ-3278 Catch That Cow(广搜+剪枝)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39887		Accepted: 12412

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

思路：

        三个方向（-1、 +1， *2 ）的广度优先搜索；都说是水题，但WA了很多次，原因在于剪枝；

        出错数据和出错代码如下：

        



        



        当k=99999时，cur.k * 2 > 99999的数据是会被剪掉的，所以100000根本没有进入队列，也就不能直接-1得出结果；

        发现问题后第一感觉就是把这条剪枝语句去掉，然后跑出23就提交了。。接着就Runtime Error了!

        这次是visit数组溢出，考虑太不全面了。。 如果题目极限数据时1W，显然去掉这句后，接近或超过一万时*2入队再以此类推就相当大了，所以这儿的剪枝还必须有的；但如何剪呢?

        关键时候大神提醒了下， cur.k * 2 <= k + 1 就行了， 原因在于， k + 2 及以后的可以由cur.k = (k + 2) / 2 时直接*2取得，就不用到这一步减1了。。想想确实精辟~

代码：

#include <stdio.h>
#include <queue>
#include <string.h>
#define N 200030

using namespace std;

struct Node{
int k;
int s;
};

int n, k;
bool visit
;
int bfs();
queue<Node>q;

int main()
{
Node nw;
while(scanf("%d%d", &n, &k)!=EOF) {
while(!q.empty()) q.pop();
memset(visit, 0, sizeof(visit));
if(n == k)
printf("0\n");
else{
nw.k = n;
nw.s = 0;
visit
= 1;

q.push(nw);
int ans = bfs();
printf("%d\n", ans);
}
}

return 0;
}

int bfs()
{
Node cur, nw;
while(1){
cur.k = q.front().k;
cur.s = q.front().s + 1;
q.pop();

if(cur.k - 1 >= 0 && !visit[cur.k - 1]){
nw = cur;
nw.k -= 1;
if(nw.k == k)
return nw.s;
visit[nw.k] = 1;
q.push(nw);
}

if(cur.k + 1 <= k && !visit[cur.k + 1]){
nw = cur;
nw.k += 1;
if(nw.k == k)
return nw.s;
visit[nw.k] = 1;
q.push(nw);
}

if(cur.k * 2 <= k && !visit[cur.k * 2]){
nw = cur;
nw.k *= 2;
if(nw.k == k)
return nw.s;
visit[nw.k] = 1;
q.push(nw);
}
}

return 0;
}