• 2019牛客暑期多校训练营(第十场)F.Popping Balloons(线段树)


    题意:现在给你n个点 现在让你横着划三条线间距为r 然后竖着划三条线间距同样为r 现在让你求经过最多的点数

    思路:我们首先建一棵关于y区间的线段树 然后枚举x轴 每次更新重叠的点 然后再更新回去 找一个最大值

    #include <bits/stdc++.h>
    using namespace std;
    const double pi = acos(-1.0);
    const int N = 1e5+7;
    const int inf = 0x3f3f3f3f;
    const double eps = 1e-6;
    typedef long long ll;
    const ll mod = 1e7+9;
    int n,d;
    struct tree{
        int l,r;
        ll v;
    }t[N<<4];
    int yy[N<<2];
    vector<int> xx[N<<2];
    void pushup(int p){
        t[p].v=max(t[p<<1].v,t[p<<1|1].v);
    }
    void build(int p,int l,int r){
        t[p].l=l; t[p].r=r;
        if(l==r){
            t[p].v=yy[l]+yy[l+d]+yy[l+2*d];
            return ;
        }
        int mid=(l+r)>>1;
        build(p<<1,l,mid);
        build(p<<1|1,mid+1,r);
        pushup(p);
    }
    void update(int p,int x,int val){
        if(t[p].l==t[p].r&&t[p].l==x){
            t[p].v+=val;
            return ;
        }
        int mid=(t[p].l+t[p].r)>>1;
        if(x<=mid) update(p<<1,x,val);
        else update(p<<1|1,x,val);
        pushup(p);
    }
    void work(int x,int val){
        update(1,x,val);
        if(x-d>=0) update(1,x-d,val);
        if(x-2*d>=0) update(1,x-2*d,val);
    }
    int main(){
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cin>>n>>d;
        int pox=0; int poy=0;
        for(int i=1;i<=n;i++){
            int x,y; cin>>x>>y;
            pox=max(pox,x); poy=max(poy,y);
            yy[y]++;
            xx[x].push_back(y);
        }
        build(1,0,poy);
        ll ans=0;
        for(int i=0;i<=pox;i++){
            ll tmp=xx[i].size()+xx[i+d].size()+xx[i+2*d].size();
            for(int j=0;j<xx[i].size();j++){
                work(xx[i][j],-1);
            }
            for(int j=0;j<xx[i+d].size();j++){
                work(xx[i+d][j],-1);
            }
            for(int j=0;j<xx[i+2*d].size();j++){
                work(xx[i+2*d][j],-1);
            }
            ans=max(ans,t[1].v+tmp);
            for(int j=0;j<xx[i].size();j++){
                work(xx[i][j],1);
            }
            for(int j=0;j<xx[i+d].size();j++){
                work(xx[i+d][j],1);
            }
            for(int j=0;j<xx[i+2*d].size();j++){
                work(xx[i+2*d][j],1);
            }
        }
        cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/wmj6/p/11372102.html
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