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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 49988 | Accepted: 15679 |
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：实在鬼畜...John在0~100，000的某个位置上，牛也在某个位置上问，John通过题述的+1，-1，*2这几种变换方式能最少变换几次捉到牛。
求最少变换，用bfs，注意标记到达的数就不能再到达，否则会结束不了循环，而且有几个剪枝：
1.John不需要跑到负数的地方，因为跑到负数显然是由-1造成的，下一步只能+1，就等于走了两步废棋。
2.John不需要跑到100，000以外的地方，因为跑到100，000以外肯定是打算*2后再减去几次到达牛的位置。但是牛在100，000以内，*2后到达的位置一定是偶数，
所以离100，000最近的偶数是100002，所以还要减2步，所以一共走了3步，所以John从50，001的位置不需*2，直接-1，再*2也到达了100，000位置，所以又多走了一步废棋，若*2后不是100002，离100，000更远那么就多走了更多废棋。
//916K 141MS
#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
queue<pair<int,int> >que;//记录到达的位置和移动的步数
int n,m;
bool book[100100];
void bfs()
{
que.push(make_pair(n,0));
book[n]=1;
while(!que.empty())
{
pair<int,int> t=que.front();
que.pop();
if(t.first==m) {printf("%d\n",t.second); break;}
if(t.first+1<=100000&&!book[t.first+1]) {que.push(make_pair(t.first+1,t.second+1));book[t.first+1]=1;}
if(t.first-1>=0&&!book[t.first-1]) {que.push(make_pair(t.first-1,t.second+1));book[t.first-1]=1;}
if(t.first*2<=100000&&!book[t.first*2]) {que.push(make_pair(t.first*2,t.second+1));book[t.first*2]=1;}
}
}
int main()
{
scanf("%d%d",&n,&m);
bfs();
return 0;
}
POJ3278Catch That Cow（线性模型）(BFS)
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原文地址：http://blog.csdn.net/kalilili/article/details/42975125
