poj_3278_Catch That Cow广搜

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 68714		Accepted: 21637

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

/*****************************************
第一次做广搜和深搜的题的时候都是一次AC,真是太高兴了（虽然是水题）
调程序的时候发现忘了写  q.pop();  真是。。。
*****************************************/

<span style="font-size:14px;">#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <bitset>
#include <list>
using namespace std;

int N,K;
int floorn[100001];
int visit[100001];
queue <int> q;

int main()
{
cin >> N >> K;                              //输入及初始化
q.push(N);
memset(visit, 0, sizeof(visit));
memset(floorn, 0, sizeof(floorn));
while(!q.empty())                           //每次取队列头节点，分层并且标记
{
int a = q.front();
visit[a] = 1;
if (a == K)                             //搜索结束条件
{
printf("%d\n",floorn[a]);
break;
}
if ((a-1 >= 0) && (!visit[a-1]))
{
q.push(a-1);
floorn[a-1] = floorn[a] + 1;
visit[a-1] = 1;
}
if ((a+1 <= 100000) && (!visit[a+1]))
{
q.push(a+1);
floorn[a+1] = floorn[a] + 1;
visit[a+1] = 1;
}
if ((2*a <= 100000) && (!visit[2*a]))
{
q.push(2*a);
floorn[2*a] = floorn[a] + 1;
visit[2*a] = 1;
}
q.pop();                                //每次头节点分析完后一定要踢掉，感觉就像深搜的回溯
}
return 0;
}</span>

一会换一种解法，再来更新

/××××××××××××××××××××××××××
好吧这次做得太慢了，因为找错找了好久没找到，
最后发现是visit[100000];没有包含坐标为100000的值。。。
失误失误。。。以后绝不会再犯这样的错


××××××××××××××××××××××××××/
<span style="font-size:14px;">#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <bitset>
#include <list>
using namespace std;

int N,K;
int visit[100010];

struct abc
{
int x;
int floorn;
abc(int aa,int bb):x(aa),floorn(bb){}
};
queue <abc> q;

int main()
{
cin >> N >> K;
q.push(abc(N,0));
memset(visit,0,sizeof(visit));
visit
= 1;
while (!q.empty())
{
abc a = q.front();
if (a.x == K)
{
printf("%d\n",a.floorn);
break;
}
if ((a.x-1 >= 0) && (!visit[a.x-1]))
{
q.push(abc(a.x-1,a.floorn+1));
visit[a.x-1] = 1;
}
if ((a.x+1 <= 100000) && (!visit[a.x+1]))
{
q.push(abc(a.x+1,a.floorn+1));
visit[a.x+1] = 1;
}
if ((2*a.x <= 100000) && (!visit[2*a.x]))
{
q.push(abc(2*a.x,a.floorn+1));
visit[2*a.x] = 1;
}
q.pop();
}
return 0;
}</span>

感谢阅读
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