如何解决POJ 3278 捕获那头牛的问题？
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本文共计657个文字，预计阅读时间需要3分钟。
题目：抓捕逃逸母牛 - 时间限制：2000MS - 内存限制：65536K - 总提交数：44220 - 已接受数：13814描述：农夫约翰得知了一头逃逸母牛的位置，他想要立即抓住它。他从点N（0
Catch That Cow
Time Limit:2000MS Memory Limit:65536K Total Submissions:44220 Accepted:13814 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXto the pointsX- 1 orX+ 1 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?
* Teleporting: FJ can move from any pointXto the point 2 ×Xin a single minute.
Input
Line 1: Two space-separated integers: Nand 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
Hint
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.
Source
USACO 2007 Open Silver
大致题意：
给定两个整数n和k
通过n+1或n-1或n*2这3种操作，使得n==k
输出最少的操作次数
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct node
{
int x,ans;
}q[1000005];
int jx[]={-1,1};
int n,k;
int v[1000005];
void BFS()
{
struct node t,f;
int i;
int e=0,s=0;
t.x=n;
v[t.x]=1;
t.ans=0;
q[e++]=t;
while(s<e)
{
t=q[s++];
if(t.x==k)
{
printf("%d\n",t.ans);
break;
}
for(i=0;i<3;i++)
{
if(i==2)
f.x=t.x*2;
else f.x=t.x+jx[i];
if(v[f.x]==0&&f.x>=0&&f.x<=100000)
{
f.ans=t.ans+1;
q[e++]=f;
v[f.x]=1;
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(v,0,sizeof(v));
BFS();
}
return 0;
}
本文共计657个文字，预计阅读时间需要3分钟。
题目：抓捕逃逸母牛 - 时间限制：2000MS - 内存限制：65536K - 总提交数：44220 - 已接受数：13814描述：农夫约翰得知了一头逃逸母牛的位置，他想要立即抓住它。他从点N（0
Catch That Cow
Time Limit:2000MS Memory Limit:65536K Total Submissions:44220 Accepted:13814 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXto the pointsX- 1 orX+ 1 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?
* Teleporting: FJ can move from any pointXto the point 2 ×Xin a single minute.
Input
Line 1: Two space-separated integers: Nand 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
Hint
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.
Source
USACO 2007 Open Silver
大致题意：
给定两个整数n和k
通过n+1或n-1或n*2这3种操作，使得n==k
输出最少的操作次数
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct node
{
int x,ans;
}q[1000005];
int jx[]={-1,1};
int n,k;
int v[1000005];
void BFS()
{
struct node t,f;
int i;
int e=0,s=0;
t.x=n;
v[t.x]=1;
t.ans=0;
q[e++]=t;
while(s<e)
{
t=q[s++];
if(t.x==k)
{
printf("%d\n",t.ans);
break;
}
for(i=0;i<3;i++)
{
if(i==2)
f.x=t.x*2;
else f.x=t.x+jx[i];
if(v[f.x]==0&&f.x>=0&&f.x<=100000)
{
f.ans=t.ans+1;
q[e++]=f;
v[f.x]=1;
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(v,0,sizeof(v));
BFS();
}
return 0;
}