HDU 2717||POJ 3278 BFS入门（理解原理）

HDU 2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11677    Accepted Submission(s): 3617

Problem 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

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

 

Recommend

teddy   |   We have carefully selected several similar problems for you:   2102  1372  1240  1072  1180 

DFS入门题目，但数据量较大。注意开数组的大小，容易RE。

北大郭炜老师课件中的源码：

#include 
#include 
#include 
using namespace std;
int N,K;
const int MAXN = 100000;
int visited[MAXN+10];  //判重标记,visited[i] = true表示i已经扩展过
struct Step
{
    int x;  //位置
    int steps;  //到达x所需的步数
    Step(int xx,int s):x(xx),steps(s) { }
};
queue q;  //队列,即Open表
int main()
{
    cin >> N >> K;
    memset(visited,0,sizeof(visited));
    q.push(Step(N,0));
    visited[N] = 1;
    while(!q.empty())
    {
        Step s = q.front();
        if( s.x == K )   //找到目标
        {
            cout << s.steps <= 0 && !visited[s.x-1] )
            {
                q.push(Step(s.x-1,s.steps+1));
                visited[s.x-1] = 1;
            }
            if( s.x + 1 <= MAXN && !visited[s.x+1] )
            {
                q.push(Step(s.x+1,s.steps+1));
                visited[s.x+1] = 1;
            }
            if( s.x * 2 <= MAXN &&!visited[s.x*2]  )
            {
                q.push(Step(s.x*2,s.steps+1));
                visited[s.x*2]  = 1;
            }
            q.pop();
        }
    }
    return 0;
}

自己写的代码：

#include 
#include 
#include 

using namespace std;

int main()
{
    int s,e;
    while(cin>>s>>e){
        queue a;
        bool y[100006];
        memset(y,0,sizeof(y));
        if(s>=e)
        {
            cout<=0&&!y[x-1]){
                if(x-1==e){cout<


参考自： 点击打开链接

把队列转化为数组：

#include 
#include 
#include 

using namespace std;

int main(){
    int s,e;
    int a[200005];//队列
    int b[200005];//记录距离

    while(cin>>s>>e){
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        int x=0;
        int i=0;
        a[0]=s;
        b[s]=1;
        while(1){
            if(a[i]==e)
            {
                cout<=0&&!b[a[i]-1]){
                a[++x]=a[i]-1;b[a[i]-1]=b[a[i]]+1;
            }
            i++;
        }
    }

}