poj--3278
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 65426 | Accepted: 20554 |
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.
解体思路：简单的bfs。
代码如下：
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int n,k,ans;
struct stu{
int x,t;
};
stu node[200000];
int vist[200000];
bool check(int x){
if(vist[x]==1)return false;
if(x<0)return false;
if(x>=200000)return false;
return true;
}
void bfs(int n){
queue<stu >q;
while(!q.empty()){
q.pop();
}
stu temp,star;
temp.x=n;temp.t=0;
q.push(temp);
while(!q.empty()){
star=q.front();
q.pop();
for(int i=1;i<=3;i++){
if(i==1){
temp.x=star.x+1;
temp.t=star.t+1;
}
else if(i==2){
temp.x=star.x-1;
temp.t=star.t+1;
}
else if(i==3){
temp.x=star.x*2;
temp.t=star.t+1;
}
if(check(temp.x)){
if(temp.x==k){
ans=temp.t;
return ;
}
q.push(temp);
vist[temp.x]=1;
}
}
}
}
int main(){
while(scanf("%d%d",&n,&k)!=EOF){
memset(vist,0,sizeof(vist));
vist[n]=1;
bfs(n);
printf("%d\n",ans);
}
return 0;
}