(离散化 + 线段树) poj 3277 City Horizon

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#tree #insert #bi #query #build #struct

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本文介绍了一种使用线段树解决城市建筑天际线总面积计算的问题，通过离散化处理大量坐标数据，有效地减少了空间消耗并提高了计算效率。

City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10318		Accepted: 2704

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

Sample Output

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.

线段树的概念：参考<<线段树在信息学竞赛中的应用>>武汉大学和<<线段树>>

/*
题目大意：在一条水平线上有N个建筑物，建筑物都是长方形的，且可以互相遮盖。给出每个建筑物的左右坐标值Ai,Bi以及每个建筑物的高度Hi，
		：需要计算出这些建筑物总共覆盖的面积。
题目解答：由于Ai，Bi∈[1,1000000000]，直接将线段[Ai,Bi]插入线段树空间消耗太大，
		：可以将这n条线段的2n个端点离散化到一个数组X[0...2n-1]，

Source Code

Problem: 3277  User: wawadimu 
Memory: 4188K  Time: 282MS 
Language: C++  Result: Accepted

Source Code
*/ 
#include<iostream>
#include<algorithm>
using namespace std;

#define maxn 40010

struct Node  //区间[l,r]内 h = 房子的最高值
{
	int l,r,h;
}tree[7*maxn];//开小了就运行出错，不知为什么3倍不行

struct XX   //原始数据(左右坐标，高度)
{
	int ai,bi,h;
}res[maxn];

int t[2*maxn];//原始坐标
int hash[maxn];//离散化坐标--空间压缩
void built(int num,int L ,int R);//建立线段树
void insert(int num,int L ,int R,int H);//维护线段树
int query(int L, int R ,int t);//查找离散化的区间
long long  Count(int num);//统计面积

int cmp ( XX a, XX b)//按告诉从小到大排列
{
	return a.h < b.h;
}

int main()
{
	//freopen("3277.txt","r",stdin);
	int n , len , i , j , k ;
	int ai , bi , h ;
	int cnt;
	while(scanf("%d",&n)!=EOF)
	{
		k=0;
		for(i=0; i<n; i++)
		{
			scanf("%d%d%d",&ai,&bi,&h);
			t[k++]=ai; t[k++]=bi;
			res[i].ai=ai; res[i].bi=bi; res[i].h=h;
		}
		sort( t , t+k );//排序
		sort( res, res+ n ,  cmp);//高度排序
		cnt=0;
		hash[cnt++]=t[0];
		for(i=1 ; i<k ;i++)//离散化
		{
			if(t[i] != t[i-1]) hash[cnt++]=t[i];
		}
		cnt--;
		//for(i=1;i<=cnt;i++)   cout<<hash[i]<<" ";cout<<endl;
		built(1,0,cnt);
		for(i=0; i <n ; i++)
		{
			int l=query(0, cnt , res[i].ai),
				r=query(l, cnt , res[i].bi);
			insert(1,l,r,res[i].h);//更新区间[l,r]的数据 h
		}
		long long  ans=Count(1);
		printf("%lld/n",ans);

}
	return 0;
}
void built(int num,int L, int R)
{

tree[num].l=L ; tree[num].r=R;//cout<<L<<":"<<R<<"  ";
	tree[num].h=-1;//区间未覆盖
	if(L+1 == R) {  return ;}
	else
	{
		int M=(L+R)>>1;
		built(2*num,L,M);
		built(2*num+1,M,R);
	}
}
void insert(int num, int L, int R ,int H)
{
	//cout<<L<<":"<<R<<"="<<H<<"->"<<tree[num].l<<tree[num].r<<" "<<tree[num].h<<"   ";
	if(H <= tree[num].h) return;
	if(L==tree[num].l && R==tree[num].r ) { tree[num].h=H; return;}
	if(tree[num].h >=0 && tree[num].l +1 != tree[num].r)
	{    tree[2*num].h=tree[2*num+1].h=tree[num].h;  tree[num].h=-1; }
	int M=(tree[num].l+tree[num].r)>>1;
	if( R <= M)      {  insert( 2*num   , L , R , H ); }
	else if( L >=M ) {  insert( 2*num+1 , L , R , H ); }
	else             {  insert( 2*num   , L , M , H );
						insert( 2*num+1 , M , R , H ); }
}
int query(int L, int R ,int d)
{
	int mid,l=L,r=R;
	while( l<= r)
	{
		mid=(l+r)>>1;
		if( d==hash[mid] ) return mid;
		else if(hash[mid] > d) r=mid-1;
		else l=mid+1;
	}
	return -1;
}
long long Count(int num)
{
	if(tree[num].l +1 == tree[num].r && tree[num].h < 0) return 0;
	if(tree[num].h >0) return tree[num].h*(long long )(hash[tree[num].r]-hash[tree[num].l]);
	else return Count(2*num) + Count(2*num+1);
}

/*
网上代码
Source Code

Problem: 3277  User: wawadimu 
Memory: 5436K  Time: 219MS 
Language: C++  Result: Accepted

Source Code
*/ 
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int A[40001],B[40001],H[40001],N=0,i,j=0,n=0,low,high,mid,LS[80002]={0},S[80001]={0};
struct{
   int l,r,h,m;
}tree[300000]={0};
void build(int p,int l,int r){
     tree[p].l=l;tree[p].r=r;
     if(r-l>1){
        int m=(l+r)>>1;
        tree[p].m=m;
        build(2*p,l,m);
        build(2*p+1,m,r);
     }
}
void insert(int p,int a,int b){
     if(S[tree[p].l]==a&&S[tree[p].r]==b){
        if(tree[p].h<H[i])tree[p].h=H[i];
        return;
     }
     int tmp=S[tree[p].m];
     if(b<=tmp)insert(2*p,a,b);
     else if(a>=tmp)insert(2*p+1,a,b);
     else{
          insert(2*p,a,tmp);
          insert(2*p+1,tmp,b);
         }
}
long long Count(int p,int h){
     if(h>tree[p].h)tree[p].h=h;
     if(tree[p].r-tree[p].l==1)return (long long)(S[tree[p].r]-S[tree[p].l])*tree[p].h;
     return Count(2*p,tree[p].h)+Count(2*p+1,tree[p].h);
}
int main(){
    //freopen("3277.txt","r",stdin);
  scanf("%d",&n);
  for(i=1;i<=n;i++){
     scanf("%d%d%d",&A[i],&B[i],&H[i]);
     LS[++j]=A[i];
     LS[++j]=B[i];
  }
  sort(LS+1,LS+1+j);
  for(i=1;i<=j;i++){
     S[++N]=LS[i];
     while(LS[i+1]==LS[i])i++;
  }
  build(1,1,N);
  for(i=1;i<=n;i++)
     insert(1,A[i],B[i]);
  cout<<Count(1,0)<<endl;
  return 0;
}