poj 3278 暴力BFS

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 63977 Accepted: 20085

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.

题意：

在一个数轴上，给两个点N，K，问N最少经过几次操作可以到K（操作可以是向后走一步，向前走一步，或向前跳淘到2x处，x为当前位置，问最少几次可以到K）

解法：

直接BFS

<iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

typedef struct{
int v,h;
}Po;

//queue<Po>qu;
int n,k;
int p[200010];

int main()
{
//    freopen("data.out","w",stdout);
while(scanf("%d%d",&n,&k)!=EOF){
queue<Po>qu;
memset(p,0,sizeof(p));
Po S;
S.h = 0;
S.v = n;
qu.push(S);
Po te;
while(!qu.empty()){
te = qu.front();qu.pop();
if(te.v == k){
break;
}
if(0<=te.v + 1&&te.v+1<=100000){
Po ter = te;
ter.v++;
ter.h++;
if(!p[ter.v]){
qu.push(ter);
p[ter.v] = 1;
}

}
if(0<=te.v - 1&&te.v - 1<=100000){
Po ter = te;
ter.v--;
ter.h++;
if(!p[ter.v]){
qu.push(ter);
p[ter.v] = 1;
}
}
if(0<=te.v*2&&te.v*2<=100000){
Po ter = te;
ter.v*=2;
ter.h++;
if(!p[ter.v]){
qu.push(ter);
p[ter.v] = 1;
}
}
}
printf("%d\n",te.h);
}
return 0;
}