# 【BZOJ1645】[Usaco2007 Open]City Horizon 城市地平线

## 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.

N个矩形块，交求面积并.

## 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

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

## Sample Output

16

OUTPUT DETAILS:

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.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define lson x<<1
#define rson x<<1|1
using namespace std;
const int maxn=40010;
typedef long long ll;
struct edges
{
ll pos,org;
}s[maxn<<1];
struct house
{
ll sa,sb,sh;
}p[maxn];
int n,m;
ll ts[maxn<<3],tag[maxn<<3],ref[maxn<<1];
bool cmp1(edges a,edges b)
{
return a.pos<b.pos;
}
bool cmp2(house a,house b)
{
return a.sh<b.sh;
}
void pushdown(int l,int r,int x)
{
if(tag[x])
{
int mid=l+r>>1;
ts[lson]=(ref[mid]-ref[l])*tag[x],ts[rson]=(ref[r]-ref[mid])*tag[x];
tag[lson]=tag[rson]=tag[x];
tag[x]=0;
}
}
void pushup(int x)
{
ts[x]=ts[lson]+ts[rson];
}
void updata(int l,int r,int x,int a,int b,ll v)
{
if(a<=l&&r<=b)
{
ts[x]=(ref[r]-ref[l])*v;
tag[x]=v;
return ;
}
pushdown(l,r,x);
int mid=l+r>>1;
if(b<=mid)    updata(l,mid,lson,a,b,v);
else if(a>=mid)    updata(mid,r,rson,a,b,v);
else    updata(l,mid,lson,a,b,v),updata(mid,r,rson,a,b,v);
pushup(x);
}
int main()
{
scanf("%d",&n);
int i;
for(i=1;i<=n;i++)
{
scanf("%lld%lld%lld",&p[i].sa,&p[i].sb,&p[i].sh);
s[i*2-1].pos=p[i].sa,s[i*2].pos=p[i].sb;
s[i*2-1].org=s[i*2].org=i;
}
sort(s+1,s+2*n+1,cmp1);
for(i=1;i<=2*n;i++)
{
if(s[i].pos>ref[m])    ref[++m]=s[i].pos;
if(s[i].pos==p[s[i].org].sa)    p[s[i].org].sa=m;
else    p[s[i].org].sb=m;
}
sort(p+1,p+n+1,cmp2);
for(i=1;i<=n;i++)
updata(1,m,1,p[i].sa,p[i].sb,p[i].sh);
printf("%lld",ts[1]);
return 0;
}

posted @ 2017-01-13 10:23  CQzhangyu  阅读(544)  评论(0编辑  收藏  举报