POJ3278Catch That Cow
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.
一个简单的bfs，就是其实本来不想写的题，但是！！
它re了
就要写一下它为什么re了
（蟹蟹师哥哥给debug）
RE代码
#include<cstdio>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define PI acos(-1)
#define inf 0x3f3f3f3f
#define EPS 1e-6
#define mem(a, b) memset(a, b, sizeof(a))
#define ll long long
#define mian main
bool visit[100010];
int dir[2] = {1, -1};
int flag;
struct node
{
int x;
int s;
};
queue<node> q;
int n, k;
int bfs()
{
while(!q.empty())
{
node m=q.front();
//cout<<m.x<<' '<<m.s<<endl;
q.pop();
//visit[m.x] = 1;
if(m.x == k)
{
return m.s;
}
node m1, m2;
for(int i = 0; i < 2; i ++)
{
m1.x = m.x + dir[i];
m1.s = m.s + 1;
if(m1.x >= 0 && m1.x <= 100000 && !visit[m1.x])
{
q.push(m1);
visit[m1.x] = 1;
}
}
m2.x = m.x * 2;
m2.s = m.s + 1;
if(!visit[m2.x] && m2.x >= 0 && m2.x <= 100000)
{
q.push(m2);
visit[m2.x] = 1;
}
}
//return 0;
}
int main()
{
while(cin>>n>>k)
{
mem(visit, 0);
while(!q.empty())
{
q.pop();
}
node pre;
visit[n] = 1;
pre.x = n;
pre.s = 0;
q.push(pre);
int ans = bfs();
cout<<ans<<endl;
}
return 0;
}AC代码
#include<cstdio>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define PI acos(-1)
#define inf 0x3f3f3f3f
#define EPS 1e-6
#define mem(a, b) memset(a, b, sizeof(a))
#define ll long long
#define mian main
bool visit[100010];
int dir[2] = {1, -1};
int flag;
struct node
{
int x;
int s;
};
queue<node> q;
int n, k;
int bfs()
{
while(!q.empty())
{
node m=q.front();
//cout<<m.x<<' '<<m.s<<endl;
//
q.pop();
//visit[m.x] = 1;
if(m.x == k)
{
return m.s;
}
node m1, m2;
for(int i = 0; i < 2; i ++)
{
m1.x = m.x + dir[i];
m1.s = m.s + 1;
if(m1.x >= 0 && m1.x <= 100000 && !visit[m1.x])
{
q.push(m1);
visit[m1.x] = 1;
}
}
m2.x = m.x * 2;
m2.s = m.s + 1;
if(m2.x >= 0 && m2.x <= 100000 && !visit[m2.x])
{
q.push(m2);
visit[m2.x] = 1;
}
}
//return 0;
}
int main()
{
while(cin>>n>>k)
{
mem(visit, 0);
while(!q.empty())
{
q.pop();
}
node pre;
visit[n] = 1;
pre.x = n;
pre.s = 0;
q.push(pre);
int ans = bfs();
cout<<ans<<endl;
}
return 0;
}你看你看，这两个有什么区别
哇啊
就第56行那个if条件里面，哇啊就应该先判断他这个有没有越界再判断有没有visit
哇啊啊