POJ 3277 City Horizon

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简单的Lazy操作。统计的时候把全部的lazy都推到叶节点就能够了

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 A_i through B_i along the horizon (1 ≤ A_i < B_i ≤ 1,000,000,000) and has height H_i (1 ≤ H_i ≤ 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:  A_i,  B_i, and  H_i

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;
}