Chef and Deforestation CodeChef - DEFOREST

https://www.codechef.com/problems/DEFOREST

对于在xi处 高度为hi的一棵树 令hi-=xi 将问题转换为求h相同的两点之间有多少点的值大于h

比如减完x后 高为h的树有p1 p2 p3...pk这几棵 在p1于p3之间连绳子相当于在p1p2之间 p2p3之间连绳子的加值之和 所以对于同一高度的树 就是个最大子段和问题 至于求h相同的两点之间有多少点的值大于h 主席树搞一搞

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double N=1000000000000000000.0;
const int maxn=250010;

struct node1
{
    ll x,h;
};

struct node2
{
    int l,r;
    ll val,cnt;
};

vector <int> pre[maxn];
node1 ary[maxn];
node2 tree[20*maxn];
double gou[maxn];
ll tmp[maxn];
int root[maxn];
int n,len,num;

bool cmp(node1 n1,node1 n2)
{
    return n1.x<n2.x;
}

int build(int l,int r)
{
    int cur,m;
    cur=num++;
    tree[cur].l=tree[cur].r=0;
    tree[cur].val=tree[cur].cnt=0;
    if(l==r) return cur;
    m=(l+r)/2;
    tree[cur].l=build(l,m);
    tree[cur].r=build(m+1,r);
    return cur;
}

int update(int rot,int tar,ll val,int l,int r)
{
    int cur,m;
    cur=num++;
    tree[cur]=tree[rot];
    tree[cur].val+=val,tree[cur].cnt++;
    if(l==r) return cur;
    m=(l+r)/2;
    if(tar<=m) tree[cur].l=update(tree[rot].l,tar,val,l,m);
    else tree[cur].r=update(tree[rot].r,tar,val,m+1,r);
    return cur;
}

void query(int lrot,int rrot,ll &res1,ll &res2,int pl,int pr,int l,int r)
{
    int m;
    if(pl<=l&&r<=pr)
    {
        res1+=(tree[rrot].val-tree[lrot].val);
        res2+=(tree[rrot].cnt-tree[lrot].cnt);
        return;
    }
    m=(l+r)/2;
    if(pl<=m) query(tree[lrot].l,tree[rrot].l,res1,res2,pl,pr,l,m);
    if(pr>m) query(tree[lrot].r,tree[rrot].r,res1,res2,pl,pr,m+1,r);
}

int main()
{
    double c,ans,minn,t1,t2;
    ll res1,res2;
    int t,i,j,p;
    c=sqrt(2.0);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%lld%lld",&ary[i].x,&ary[i].h);//sort
            ary[i].h-=ary[i].x;
            //tmp[i]=h[i];
        }
        sort(ary+1,ary+n+1,cmp);
        for(i=1;i<=n;i++) tmp[i]=ary[i].h;
        /*
        for(i=1;i<=n;i++) printf("%lld ",h[i]);
        printf("\n");
        */
        sort(tmp+1,tmp+n+1);
        len=unique(tmp+1,tmp+n+1)-tmp-1;
        for(i=1;i<=len;i++) pre[i].clear();
        num=0;
        root[0]=build(1,len);
        for(i=1;i<=n;i++)
        {
            p=lower_bound(tmp+1,tmp+len+1,ary[i].h)-tmp;
            pre[p].push_back(i);
            root[i]=update(root[i-1],p,ary[i].h,1,len);
        }
        /*
        for(i=1;i<=len;i++)
        {
            printf("**")
        }
        */
        ans=-N;
        for(i=1;i<=len;i++)
        {
            for(j=0;j+1<pre[i].size();j++)
            {
                t1=ary[pre[i][j+1]].x-ary[pre[i][j]].x;
                t1*=c;
                p=lower_bound(tmp+1,tmp+len+1,ary[pre[i][j]].h)-tmp;
                if(p+1<=len)
                {
                    res1=0,res2=0;
                    query(root[pre[i][j]-1],root[pre[i][j+1]],res1,res2,p+1,len,1,len);
                    t2=res1-res2*tmp[p];
                }
                else t2=0.0;
                gou[j]=t1-t2;
            }
            minn=0.0;
            for(j=0;j+1<pre[i].size();j++)
            {
                if(j>0) gou[j]+=gou[j-1];
                ans=max(ans,gou[j]-minn);
                /*
                if(ans<gou[j]-minn)
                {
                    printf("%d %d\n",pre[i][j],pre[i][j+1]);
                    ans=gou[j]-minn;
                }
                */
                minn=min(minn,gou[j]);
            }
        }
        if(ans!=-N) printf("%.8f\n",ans);
        else printf("-1\n");
    }
    return 0;
}

/*
1
3
1 1
2 5
3 3

1
6
2 1
4 11
3 10
6 10
5 4
7 6

1
4
2 1
6 10
5 4
7 6
*/