poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 74854 Accepted: 23630

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

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【分析】
最近迷之懒惰，不想更博，然而考虑到老师可能会查我水表，还是先写上两篇。
题意：男主起初在n点，每步可以走前一个位置，后一个位置，或者二倍的位置，问最少几步能到达指定点。
那就宽搜一下出解（标记过的点就不走了，别人说要剪枝什么的其实不用），复杂度   O（n）
see you

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【代码】

//poj 3278 Catch That Cow#include<iostream>#include<cstdio>#include<cstring>#define fo(i,j,k) for(i=j;i<=k;i++)#define M(a) memset(a,0,sizeof a)using namespace std;int n,m;int q[200001],map[200001];bool vis[200001];int main(){    int i,j,k,u,v;    scanf("%d%d",&n,&m);    if(n>m) printf("%d\n",n-m);    int h=0,t=1;    q[t]=n;    vis[n]=1;    while(h<t)    {        h++;        u=q[h];        v=u-1;        if(v>=0 && v<=m+5 && !vis[v])        {            vis[v]=1;            q[++t]=v;            map[v]=map[u]+1;        }        v=u+1;        if(v>=0 && v<=m+5 && !vis[v])        {            vis[v]=1;            q[++t]=v;            map[v]=map[u]+1;        }        v=u+u;        if(v>=0 && v<=m+5 && !vis[v])        {            vis[v]=1;            q[++t]=v;            map[v]=map[u]+1;        }        if(vis[m]) {printf("%d\n",map[m]);return 0;}    }