joj 2431: Shift and Increment (模板队列与数组模拟队列的对比练习)

最新推荐文章于 2014-09-15 18:42:00 发布

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最新推荐文章于 2014-09-15 18:42:00 发布

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文章标签： struct integer basic output input 优化

本文链接：https://blog.youkuaiyun.com/jxy859/article/details/6342606

本文介绍了如何通过广度优先搜索（BFS）解决从给定数字x到0的最短路径问题，利用位运算进行优化。包括使用STL模板和数组模拟队列的方法实现，特别关注于效率优化和技术细节。

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	5s	16384K	1226	196	Standard

Shift and increment is the basic operations of the ALU (Arithmetic Logical Unit) in CPU. One number can be transform to any other number by these operations. Your task is to find the shortest way from x to 0 using the shift (*=2) and increment (+=1) operations. All operations are restricted in 0..n, that is if the result x is greater than n, it should be replace as x%n.

Input and Output

There are two integer x, n (n <=1000000)

Sample Input

2 4
3 9

Sample Output

1
3
//两种模拟队列，模板库的是0.24S，模拟数组是0.14s，明显的优势啊。这题优化的关键是标记父亲节点，bfs题，貌似有数学方法。

//先贴STL模板

#include <cstdio>
#include <iostream>
#include <queue>
#include <memory.h>
using namespace std;
const int maxn=1000010;
struct state
{
    bool flag;
    int step;
}a[maxn];
int x,n,i;
int main ()
{
    while (scanf("%d%d",&x,&n)==2)
    {
        for(  i=0 ; i<n ; i++)
        {
            a[i].flag=1;
            a[i].step=-1;
        }
        int t;
        queue<int>q ;
        a[x].step=0;
        q.push(x);
        while (!q.empty())
        {
            int t;
            t=q.front();
            int t1=t*2%n;
            int t2=(t+1)%n;
            if(a[t].flag)
            {
                if(a[t1].flag)
                {
                    if(a[t1].step==-1)
                     a[t1].step=a[t].step+1;
                    q.push(t1);//先放入队列，若以搜索到结果则从队尾取。
                    //printf("t=%d/n",t);
                }
                    if(t1==0)break;
                if(a[t2].flag)
                {
                    if(a[t2].step==-1)
                     a[t2].step=a[t].step+1;
                    q.push(t2);
                    //printf("t+1=%d/n",(t+1)%n);
                }
                    if(t2==0)break;
            }
            a[t].flag=0;
            q.pop();
        }
        printf("%d/n",a[q.back()].step);;
    }
    return 0;
}


//数组模拟队列方法，时效大大地明显

#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;
const int maxn=1000010;
struct state
{
    bool flag;
    int step;
}a[maxn];
int q[1500000];
int x,n,i;
int main ()
{
    while (scanf("%d%d",&x,&n)==2)
    {
        for(  i=0 ; i<n ; i++)
        {
            a[i].flag=1;
            a[i].step=-1;
        }
        int t;
        int rear=1,head=0;
        a[x].step=0;
        q[head]=x;
        while (rear>=head)
        {
            int t;
            t=q[head];
            int t1=t*2%n;
            int t2=(t+1)%n;
            if(a[t].flag)
            {
                if(a[t1].flag)
                {
                    if(a[t1].step==-1)
                     a[t1].step=a[t].step+1;
                    q[rear++]=t1;//先放入队列，若以搜索到结果则从队尾取。
                    //printf("t=%d/n",t);
                    if(t1==0)break;
                }
                if(a[t2].flag)
                {
                    if(a[t2].step==-1)
                     a[t2].step=a[t].step+1;
                    q[rear++]=t2;
                    //printf("t+1=%d/n",(t+1)%n);
                    if(t2==0)break;
                }
            }
            a[t].flag=0;
            head++;
        }
        printf("%d/n",a[q[rear-1]].step);;
    }
    return 0;
}