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POJ 3277 City Horizon

2024-07-12 12:29:19   220人阅读


简单的Lazy操作，统计的时候把所有的lazy都推到叶节点就可以了

City Horizon
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 15973 Accepted: 4336

Description

Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings.

The entire horizon is represented by a number line with N (1 ≤ N ≤ 40,000) buildings. Building i‘s silhouette has a base that spans locations Ai through Bi along the horizon (1 ≤ Ai < Bi ≤ 1,000,000,000) and has height Hi (1 ≤ Hi ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.

Input

Line 1: A single integer: N 
Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi

Output

Line 1: The total area, in square units, of the silhouettes formed by all N buildings

Sample Input

4
2 5 1
9 10 4
6 8 2
4 6 3

Sample Output

16

Hint

The first building overlaps with the fourth building for an area of 1 square unit, so the total area is just 3*1 + 1*4 + 2*2 + 2*3 - 1 = 16.

Source

USACO 2007 Open Silver


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long int LL;

const LL maxn=204000;

LL r[maxn*2],md[maxn*2],val[maxn*2],n,m,vt,zhi[maxn*2],Hash[maxn*2],high[maxn];

bool cmpR(int a,int b)
{
   return val[a]<val[b];
}

int Lisan(int n)
{
    for(int i=0;i<n;i++) r[i]=i;
    sort(r,r+n,cmpR);
    md[0]=val[r[0]];
    val[r[0]]=m=0;
    for(int i=1;i<n;i++)
    {
        if(md[m]!=val[r[i]])
            md[++m]=val[r[i]];
        val[r[i]]=m;
    }
    return m;
}

LL tree[maxn<<2],cover[maxn<<2];

void push_down(int l,int r,int rt)
{
    if(cover[rt])
    {
        tree[rt<<1]=max(tree[rt<<1],tree[rt]);
        tree[rt<<1|1]=max(tree[rt<<1|1],tree[rt]);
        cover[rt<<1]=cover[rt<<1|1]=1;
        cover[rt]=0;
    }
}

void update(int L,int R,LL c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        tree[rt]=max(tree[rt],c);
        cover[rt]=1;
        return ;
    }
    push_down(l,r,rt);
    int m=(l+r)/2;
    if(L<=m) update(L,R,c,lson);
    if(R>m) update(L,R,c,rson);
}

LL ans;

void over_tree(int l,int r,int rt)
{
    push_down(l,r,rt);
    if(l==r)
    {
        ans+=(Hash[r+1]-Hash[l])*tree[rt];
        return ;
    }
    int m=(l+r)/2;
    over_tree(lson); over_tree(rson);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        vt=0;
        memset(tree,0,sizeof(tree));
        memset(cover,0,sizeof(cover));
        for(int i=0;i<n;i++)
        {
            LL a,b,c;
            scanf("%I64d%I64d%I64d",&a,&b,&c);
            val[vt]=zhi[vt]=a; vt++;
            val[vt]=zhi[vt]=b; vt++;
            high[i]=c;
        }
        int mx=Lisan(vt);
        for(int i=0;i<vt;i++)
        {
            Hash[val[i]]=zhi[i];
        }
        for(int i=0;i<n;i++)
        {
            update(val[i*2],val[i*2+1]-1,high[i],0,mx,1);
        }
        ans=0;
        over_tree(0,mx,1);
        printf("%I64d\n",ans);
    }
    return 0;
}



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