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POJ-3278 Catch That Cow
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
大数组可以用vector
大数组可以用vector
大数组可以用vector
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int N, K;
vector<int> data;
int main(void)
{
int bfs(void);
while(scanf("%d%d", &N, &K) != EOF)
{
for(int i = 0; i <= 2*K; i++)
data.push_back(-1);
if(K <= N)
{
printf("%d\n", N-K);
continue;
}
else
{
int e = bfs();
printf("%d\n", e);
}
}
return 0;
}
int bfs()
{
queue<int> que;
int cas;
data[N] = 0;
cas = N;
que.push(cas);
while(que.size())
{
int ca = que.front(); que.pop();
if(ca == K)
break;
for(int i = 0; i < 3; i++)
{
if(i == 0)
cas = ca + 1;
if(i == 1)
cas = ca - 1;
if(i==2)
cas = ca * 2;
if(cas >= 0 && cas <= 2*K && data[cas] < 0)
{
data[cas] = data[ca] + 1;
que.push(cas);
}
}
}
return data[K];
}
POJ-3278 Catch That Cow
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