POJ 3277 City Horizon（扫描线）

最新推荐文章于 2019-01-08 01:52:28 发布

原创最新推荐文章于 2019-01-08 01:52:28 发布 · 634 阅读

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本文介绍了一种计算在坐标轴上排列的多个建筑群总面积的方法，包括输入数据格式、算法思路及实例解析。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17061		Accepted: 4653

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.

Source

USACO 2007 Open Silver

题意：有n做建筑在一个坐标轴上（想象在坐标轴上），三个数分别表示为建筑在坐标轴上的起点终点建筑的高度，然后求所有建筑面积的和（可能会出现两个建筑有一部分重合，这样的话重合的部分算一次就可以了）

思路：该题的做法和poj1389的思路差不多，不同的是1389是从上到下扫描，而本题是从左到右进行扫描。两个题几乎是一样的。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N  40011

using namespace std;

struct node
{
    long long int lf,rf;
    long long int cnt,ans;
    long long int l,r;
}q[N<<3];

struct tt
{
    long long int x1;
    long long int x2;
    long long int y;
    long long int flag;
}p[N<<3];

int n;
int k;
long long int pnum[N<<2];

bool cmp(tt a,tt b)
{
    if(a.y == b.y)
    {
        return a.x1 < b.x1;
    }
    return a.y<b.y;
}

void build(long long int l,long long int r,long long int rt)
{
    q[rt].l = l;
    q[rt].r = r;
    q[rt].ans = 0;
    q[rt].cnt = 0;
    q[rt].lf = pnum[l];
    q[rt].rf = pnum[r];
    if(l + 1 == r)
    {
        return ;
    }
    int mid = (l+r)>>1;
    build(l,mid,rt<<1);
    build(mid,r,rt<<1|1);
}

void updata(long long int rt)
{
    if(q[rt].ans > 0)
    {
        q[rt].cnt = q[rt].rf - q[rt].lf;
        return ;
    }
    else
    {
        q[rt].cnt = 0;
    }
    if(q[rt].l + 1  == q[rt].r)
    {
        return ;
    }
    else
    {
        q[rt].cnt = q[rt<<1].cnt + q[rt<<1|1].cnt;
    }
}

void insert(long long int rt,tt dot)
{
    if(q[rt].lf == dot.x1 && q[rt].rf == dot.x2)
    {
        q[rt].ans += dot.flag;
        updata(rt);
        return ;
    }
    if(q[rt<<1].rf>=dot.x2)
    {
        insert(rt<<1,dot);
    }
    else if(q[rt<<1|1].lf<=dot.x1)
    {
        insert(rt<<1|1,dot);
    }
    else
    {
        struct tt pt = dot;
        pt.x2 = q[rt<<1].rf;
        insert(rt<<1,pt);
        pt = dot;
        pt.x1 = q[rt<<1|1].lf;
        insert(rt<<1|1,pt);
    }
    updata(rt);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        long long int x1,x2,y;
        k = 1;
        for(int i=0;i<n;i++)
        {
            scanf("%lld%lld%lld",&x1,&x2,&y);
            p[k].x1 = x1;
            p[k].x2 = x2;
            p[k].y = 0;
            p[k].flag = 1;
            pnum[k] = x1;
            k++;
            p[k].x1 = x1;
            p[k].x2 = x2;
            p[k].y = y;
            p[k].flag = -1;
            pnum[k] = x2;
            k++;
        }
        k--;
        sort(p+1,p+k+1,cmp);
        sort(pnum+1,pnum+1+k);
        build(1,k,1);
        insert(1,p[1]);
        long long int sum = 0;
        for(int i=2;i<=k;i++)
        {
            sum += q[1].cnt * (p[i].y - p[i-1].y);
            insert(1,p[i]);
        }
        printf("%lld\n",sum);
    }
    return 0;
}