Codeforces755D (segment tree,implementation)

本文介绍了一种利用线段树模拟解决凸多边形切割问题的方法,通过追踪多边形上的路径并更新线段树来计算覆盖路径的最优解。文章详细展示了算法流程,并附带了完整的代码实现。

凸多边形切割问题:线段树模拟可以做

  • 优弧与劣弧分两种情况考虑
  • 最后结束点一定是起始点
  • 线段树tree_add需要四个参数,tree_sum需要五个参数
  • 线段树所用空间不超maxn*4
#include <cstdio>
#include <cstring>

using namespace std;

#define maxn 1000005

int node[maxn];
long long tree[maxn*4];

void tree_add(int now,int l,int r,int key)
{
    tree[now]++;
    if(l==r) return ;

    int mid = (l+r)/2;
    if(key>mid)
        tree_add(now*2+2,mid+1,r,key);
    else
        tree_add(now*2+1,l,mid,key);
}

long long tree_sum(int now,int l,int r,int keyl,int keyr)
{
    if(l==keyl&&r==keyr) return tree[now];

    int mid = (l+r)/2;
    if(keyl>mid)
        return tree_sum(now*2+2,mid+1,r,keyl,keyr);
    else if(keyr<=mid)
        return tree_sum(now*2+1,l,mid,keyl,keyr);
    else
        return tree_sum(now*2+1,l,mid,keyl,mid)+tree_sum(now*2+2,mid+1,r,mid+1,keyr);
}

int main()
{
    memset(node,0,sizeof(node));
    memset(tree,0,sizeof(tree));

    int N,k;
    scanf("%d%d",&N,&k);

    long long ans = 1;
    int loc = 0;
    int nloc = loc+k;
    node[loc] = 1;

    while(node[nloc]==0)
    {
        node[nloc] = 1;
        tree_add(0,0,N-1,loc);
        tree_add(0,0,N-1,nloc);

        ans++;
        if(nloc<loc)
        {
            if(k>N/2)
                ans+=tree_sum(0,0,N-1,nloc+1,loc-1);
            else
            {
                ans+=tree_sum(0,0,N-1,0,nloc-1);
                if(loc!=N-1)
                    ans+=tree_sum(0,0,N-1,loc+1,N-1);
            }
        }else
            if(k>N/2){
                if(loc!=0)
                    ans += tree_sum(0,0,N-1,0,loc-1);
                if(nloc!=N-1)
                    ans+=tree_sum(0,0,N-1,nloc+1,N-1);
            }else
                ans+=tree_sum(0,0,N-1,loc+1,nloc-1);

        loc = nloc;
        nloc = (nloc+k)%N;
        printf("%I64d ",ans);
    }
    ans++;
    if(k>N/2)
        ans+=tree_sum(0,0,N-1,1,loc-1);
    else
        ans+=tree_sum(0,0,N-1,loc+1,N-1);
    printf("%I64d\n",ans);
    return 0;
}

封装后代码

#include <cstdio>
#include <cstring>

#define maxn 1000005

int node[maxn];
long long tree[maxn*4];


/*
the segment gnum must full binary tree,
the left child is loc*2+1,
and the right child is loc*2+2.

#define maxn 1000005

int gnode[maxn];
long long gnum[maxn*4];
*/
void GxdSegmentTreeAdd(long long* gnum,int glen,int gfl,int gfr,int gnow,int gkeyloc)
{
    gnum[gnow]++;
    if(gfl==gfr) return ;
    int mid = (gfl+gfr)/2;
    if(gkeyloc>mid) GxdSegmentTreeAdd(gnum,glen,mid+1,gfr,gnow*2+2,gkeyloc);
    else GxdSegmentTreeAdd(gnum,glen,gfl,mid,gnow*2+1,gkeyloc);
}

long long GxdSegmentTreeSearch(long long* gnum,int glen,int gfl,int gfr,int gnow,int gkeyl,int gkeyr)
{
    if(gfl==gkeyl&&gfr==gkeyr) return tree[gnow];
    int mid = (gfl+gfr)/2;
    if(gkeyl>mid) return GxdSegmentTreeSearch(gnum,glen,mid+1,gfr,gnow*2+2,gkeyl,gkeyr);
    else if(gkeyr<=mid) return GxdSegmentTreeSearch(gnum,glen,gfl,mid,gnow*2+1,gkeyl,gkeyr);
    else return GxdSegmentTreeSearch(gnum,glen,gfl,mid,gnow*2+1,gkeyl,mid)+GxdSegmentTreeSearch(gnum,glen,mid+1,gfr,gnow*2+2,mid+1,gkeyr);
}

int main()
{
    memset(node,0,sizeof(node));
    memset(tree,0,sizeof(tree));
    int N,k;
    scanf("%d%d",&N,&k);
    long long ans = 1;
    int loc = 0;
    int nloc = loc + k;
    node[loc] = 1;

    while(node[nloc]==0)
    {
        node[nloc] = 1;
        GxdSegmentTreeAdd(tree,maxn*4,0,N-1,0,loc);
        GxdSegmentTreeAdd(tree,maxn*4,0,N-1,0,nloc);

        ans++;
        if(nloc<loc)
        {
            if(k>N/2)
                ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,nloc+1,loc-1);
            else
            {
                ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,0,nloc-1);
                if(loc!=N-1)
                    ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,loc+1,N-1);
            }
        }else
            if(k>N/2){
                if(loc!=0)
                    ans += GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,0,loc-1);
                if(nloc!=N-1)
                    ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,nloc+1,N-1);
            }else
                ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,loc+1,nloc-1);

        loc = nloc;
        nloc = (nloc+k)%N;
        printf("%I64d ",ans);
    }
    ans++;
    if(k>N/2)
        ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,1,loc-1);
    else
        ans+=GxdSegmentTreeSearch(tree,maxn*4,0,N-1,0,loc+1,N-1);
    printf("%I64d\n",ans);

    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?
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值