搜索练习6/poj.org/problem3278 /Catch That Cow/简单的bfs模板
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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
Output
Sample Input
5 17
Sample Output
4
Hint
题意很简单:一维的坐标,从一点到另一点最少要走多少步;
思路:每一步有三种可能。+1,-1,*2;简单的bfs
AC代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxx = 100100;
const int N = 105;
int n,m;
int v[maxx];
struct node {int y,step;}start,endd;
int bfs(node x,node s)
{
node p1,p2;
queue<node>q;
q.push(x);
v[x.y]=1;
while(!q.empty())
{
p1=q.front();
if(p1.y==endd.y)
return p1.step;
q.pop();
if(p1.y>0&&!v[p1.y-1])
{
p2=p1;
p2.y-=1;
p2.step++;
v[p2.y]=1;
q.push(p2);
}
if(p1.y<100000&&!v[p1.y+1])
{
p2=p1;
p2.y+=1;
v[p2.y]=1;
p2.step++;
q.push(p2);
}
if(p1.y*2<=100000&&!v[p1.y*2])
{
p2=p1;
p2.y*=2;
v[p2.y]=1;
p2.step++;
q.push(p2);
}
}
return -1;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(v,0,sizeof(v));
start.y=n,start.step=0;
endd.y=m;
printf("%d\n",bfs(start,endd));
}
}
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