poj 3278 catch the cow

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.

///数据都过了 runtime error 呢 只能先贴上 明天再解决了
#include <iostream>
#include<cstring>
#include<queue>
const int maxn = 100000;///要注意数据范围呀
struct node
{
int x;
int t;
};
node S,T;
node New,now;
bool vis[maxn+2];
int pos[maxn+2];
using namespace std;
int bfs()
{
memset(vis,false,sizeof(vis));
queue<node>q;
q.push(S);
vis[S.x] = true;
while(!q.empty())
{
now = q.front();
q.pop();
New.x = now.x + 1;///************
New.t = now.t + 1;这部分这样写也不对，因为这是并列的三种情况呀
vis[New.x] = true;
q.push(New);
if(New.x == T.x)
return New.t;
New.x = now.x - 1;
New.t = now.t + 1;
vis[New.x] = true;
q.push(New);
if(New.x == T.x)
return New.t;
New.x = now.x * 2;
New.t = now.t + 1;
vis[New.x] = true;
q.push(New);
if(New.x == T.x)
return New.t;///***************
}
return -1;///这里是不能return -1的，因为至少走0步呀
}
int main()
{
int start, ending;
cin >> start >> ending;
S.x = start;
S.t = 0;
T.x = ending;///要清空队列
int ans = bfs();///万一start >=ending 呢
if(ans > 0)
cout << ans << endl;
return 0;
}

继续解决昨天的问题

上面的解法是不对的

修改如上：

漏洞好多

<pre name="code" class="cpp">
#include <iostream>
#include<cstring>
#include<queue>
using namespace std;
struct node
{
int x;
int t;
};
int start, ending;
node S,T;
node New,now;
const int maxn = 200000;
bool vis[maxn+2];
int pos[maxn+2];
queue<node>q;
int bfs()
{
memset(vis,false,sizeof(vis));

q.push(S);
vis[S.x] = true;
while(!q.empty())
{
now = q.front();
q.pop();

/*for(int i=0; i<3; i++)
{
if(i==0)
{
New.x = now.x + 1;

}
else if(i==1)
{
New.x = now.x - 1;

}
else
{
New.x = now.x * 2;

}
New.t = now.t + 1;
if(New.x >=maxn ||New.x <0 || vis[New.x])
continue;
vis[New.x] = true;
q.push(New);
if(New.x == T.x)
return New.t;
}*/

}
return 0;
}
int main()
{

while(cin >> start >> ending)
{
S.x = start;
S.t = 0;
T.x = ending;
while(!q.empty())
q.pop();

if(start >= ending)
cout << start - ending << endl;
else
{
int ans = bfs();
if(ans > 0)
cout << ans << endl;
}
}

return 0;
}

酱紫~~ 就完美了