BZOJ 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛
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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
Farmer John starts at point 5 and the fugitive cow is at point 17.
Sample Output
OUTPUT DETAILS:
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.
题解
bfs即可。因为可以到x*2也可以往回走,但注意往回走只能一个一个走,所以bfs边界x<=n+k,因为,若x=2*n且x>n+k(n<k)则要往回走>n次才能到k,还不如一开始就往回走到k整除2的地方再一个两倍直接到k或k-1。
#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<cmath>#include<algorithm>#define inf 1<<30using namespace std;int n,m,t,w,x,b,a[200005],q[200005],mark[200005];int main(){scanf("%d%d",&n,&m);if(n==m) {printf("0\n"); return 0;}for(int i=0;i<=n+m;i++) a[i]=inf;t=0; w=1; q[0]=n; a[n]=0; mark[n]=1; while(t!=w) {x=q[t]; t=(t+1)%200005; if(x-1>=0&&a[x-1]>a[x]+1) {a[x-1]=a[x]+1; if(!mark[x-1]) {q[w]=x-1; mark[x-1]=1; w=(w+1)%200005;} }if(x+1<=n+m&&a[x+1]>a[x]+1) {a[x+1]=a[x]+1; if(!mark[x+1]) {q[w]=x+1; mark[x+1]=1; w=(w+1)%200005;} }if(x<<1<=n+m&&a[x<<1]>a[x]+1) {a[x<<1]=a[x]+1; if(!mark[x<<1]) {q[w]=x<<1; mark[x<<1]=1; w=(w+1)%200005;} }mark[x]=0;if(a[m]<inf) {printf("%d\n",a[m]); break;} }return 0;}
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