| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 44613 | Accepted: 13946 |
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
题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用BFS来搜索这三个状态，直到搜索到牛所在的位置。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 200100;
int n, k;
struct node
{
int x, step;
};
queue<node> q;
int vist[N];
void bfs()
{
int cow, ans;
while(!q.empty())
{
node tmp = q.front();
q.pop();
cow = tmp.x;
ans = tmp.step;
if(cow == k)
{
printf("%d\n",ans);
return ;
}
if(cow >= 1 && !vist[cow - 1]) //要保证减1后有意义,所以要cow >= 1 减一的情况
{
node temp;
vist[cow - 1] = 1;
temp.x = cow - 1;
temp.step = ans + 1;
q.push(temp);
}
if(cow <= k && !vist[cow + 1]) //加1的情况
{
node temp;
vist[cow + 1] = 1;
temp.x = cow + 1;
temp.step = ans + 1;
q.push(temp);
}
if(cow <= k && !vist[cow * 2]) //乘二的情况
{
node temp;
vist[cow * 2] = 1;
temp.x = 2 * cow;
temp.step = ans + 1;
q.push(temp);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
while(!q.empty()) q.pop();
memset(vist,0,sizeof(vist));
vist[n] = 1;
node t;
t.x = n, t.step = 0;
q.push(t);
bfs();
}
return 0;
}
POJ 3278： Catch That Cow,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u013487051/article/details/37594177
