poj3277 City Horizon-优快云博客

本文链接：https://blog.youkuaiyun.com/bestcode_boss/article/details/38126977

本文介绍了一种使用离散化加线段树的数据结构解决计算多个矩形建筑形成的天际线总面积问题的方法。该方法首先对建筑边界进行排序，然后通过线段树更新高度并查询总面积。

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H. City Horizon

Time Limit: 2000ms

Case Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

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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

Sample Output

一个JB题，坑了我2个小时。。。

就因为把sort(w+1,w+1+n)写成了sort（w,w+n,cmp）

离散化+线段树

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int N=80001;
struct acm
{
int lt;
int rt;
long long h;
}node[N*4];
struct wp
{
int s;
int e;
long long height;
}w[N];
int p[N];
long long ans;
void build(int l,int r,int id)
{
node[id].rt=r;
node[id].lt=l;
node[id].h=0;
if(r-l==1)
return;
int mid=(l+r)>>1;
build(l,mid,id<<1);
build(mid,r,id<<1|1);
}
void push_down(int id){
if(node[id].h){
node[id<<1].h=node[id<<1|1].h=node[id].h;
node[id].h=0;
}
}
void update(int l,int r,long long hh,int id)
{
if(node[id].rt==r&&node[id].lt==l)
{
node[id].h=hh;
return;
}
if(node[id].lt==node[id].rt-1)
return;
push_down(id);
int mid=(node[id].lt+node[id].rt)>>1;
if(r<=mid)
update(l,r,hh,id<<1);
else if(l>=mid)
update(l,r,hh,id<<1|1);
else
{
update(l,mid,hh,id<<1);
update(mid,r,hh,id<<1|1);
}
}

void query(int id)
{
if(node[id].h>0)
{
//cout<<p[node[id].rt-1]<<" "<<p[node[id].lt-1]<<" "<<node[id].h<<endl;
ans+=(p[node[id].rt-1]-p[node[id].lt-1])*node[id].h;
return;
}
if(node[id].lt+1==node[id].rt)
return;
query(id<<1);
query(id<<1|1);
}

int cmp( wp a, wp b)
{
return a.height<b.height;
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
build(1,N,1);
int cnt=0;
for(int i=1;i<=n;i++)
{
scanf("%d%d%lld",&w[i].s,&w[i].e,&w[i].height);
p[cnt++]=w[i].s;
p[cnt++]=w[i].e;
}
sort(p,p+cnt);
sort(w+1,w+n+1,cmp);
cnt=unique(p,p+cnt)-p;

for(int i=1;i<=n;i++)
{
int ll=lower_bound(p,p+cnt,w[i].s)-p+1;
int rr=lower_bound(p,p+cnt,w[i].e)-p+1;
update(ll,rr,w[i].height,1);
}
ans=0;
query(1);
printf("%lld\n",ans);
}
return 0;
}