CodeForces 46D Parking Lot (线段树区间合并)

本文介绍了一种使用线段树解决停车问题的方法。通过区间合并实现车辆的停放与离开,巧妙地处理了边界条件,确保算法高效运行。文章详细阐述了解决方案的设计思路与实现细节。

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题意:

长为len的停车场,一辆车要停下,则必须车尾留b长度的空间且车头与前面的车保持f长度。

有n次操作:1,长为val 的车要寻找车位停下,若能停下输出最小的停车点下标,否则输出-1

                    2,第val辆车开走了。

解:显然是线段树的区间合并。这里有个小技巧,就是树的范围是(-b,len+f-1).这里就保证了第一辆车停在0点,不用特判边界条件。为什么是len+f-1而不是len+f呢?因为,我的做法是这个点代表单位长度 ,0点也表示长度。

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
#define LL long long
const int INF=0x3f7f7f7f;
const int MAXN=101005;
const double eps = 1e-10;
struct node
{
    int l,r,ll,rr,mx,c;
}tr[MAXN*4];
struct point
{
    int l,r;

}a[MAXN];
void pushup(int  id);
int max(int x,int y,int z)
{
    return max(x,max(y,z));
}
void build(int id,int l,int r)
{
    tr[id].l=l;
    tr[id].r=r;
    tr[id].c=0;
    if(l==r)
    {
        tr[id].ll=tr[id].rr=tr[id].mx=1;
    }
    else
    {
        int mid=(l+r)>>1;
        build(id*2,l,mid);
        build(id*2+1,mid+1,r);
        pushup(id);
    }
}
void pushup(int id)
{
    tr[id].ll=tr[id*2].ll;
    tr[id].rr=tr[id*2+1].rr;
    if(tr[id].ll==tr[id*2].r-tr[id*2].l+1)tr[id].ll+=tr[id*2+1].ll;
    if(tr[id].rr==tr[id*2+1].r-tr[id*2+1].l+1)tr[id].rr+=tr[id*2].rr;
    tr[id].mx=max(tr[id*2].mx,tr[id*2+1].mx,tr[id*2].rr+tr[id*2+1].ll);
}
void pushdown(int id)
{
    tr[id*2].c=tr[id*2+1].c=1;
    tr[id].c=0;
    if(tr[id].mx)
    {
        tr[id*2].ll=tr[id*2].rr=tr[id*2].mx=tr[id*2].r-tr[id*2].l+1;
        tr[id*2+1].ll=tr[id*2+1].rr=tr[id*2+1].mx=tr[id*2+1].r-tr[id*2+1].l+1;
    }
    else
    {
        tr[id*2].ll=tr[id*2].rr=tr[id*2].mx=0;
        tr[id*2+1].ll=tr[id*2+1].rr=tr[id*2+1].mx=0;
    }

}
int quary(int id,int mx)
{
    if(tr[id].l==tr[id].r)
    {
        return tr[id].l;
    }
    else
    {
        //int mid=(tr[id].l+tr[id].r)/2;
        if(tr[id].c)pushdown(id);
        if(mx<=tr[id*2].mx)return quary(id*2,mx);
        else if(mx<=tr[id*2].rr+tr[id*2+1].ll)
        {
            return tr[id*2].r-tr[id*2].rr+1;
        }
        else
        {
            return quary(id*2+1,mx);
        }
    }
}
void update(int id,int l,int r,int val)
{
    if(l<=tr[id].l&&tr[id].r<=r)
    {
        tr[id].c=1;
        if(val==0)
        tr[id].ll=tr[id].rr=tr[id].mx=tr[id].r-tr[id].l+1;
        else
            tr[id].ll=tr[id].rr=tr[id].mx=0;
    }
    else
    {
        int mid=(tr[id].l+tr[id].r)>>1;
        if(tr[id].c)pushdown(id);
        if(r<=mid)update(id*2,l,r,val);
        else if(l>mid)update(id*2+1,l,r,val);
        else
        {
            update(id*2,l,r,val);
            update(id*2+1,l,r,val);

        }
        pushup(id);
    }
}
int main()
{
    int len,i,q,b,f;
    while(scanf("%d%d%d%d",&len,&b,&f,&q)!=EOF)
    {
        build(1,-b,len+f-1);
        int cnt=0;
        for(i=1;i<=q;i++)
        {
            int op,val;
            scanf("%d%d",&op,&val);
            if(op==1)
            {
                if(tr[1].mx<(val+b+f))
                {
                    puts("-1");
                    a[i].l=-1;
                    a[i].r=-1;
                    continue;
                }
                int ans=quary(1,val+b+f);
                printf("%d\n",ans+b);
                a[i].l=ans+b;
                a[i].r=ans+b+val-1;
                update(1,a[i].l,a[i].r,1);
            }
            else
            {
                a[i].l=-1;
                a[i].r=-1;
                update(1,a[val].l,a[val].r,0);
            }
        }
    }
    return 0;
}



### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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