POJ 3277 City Horizon 矩形切割

最新推荐文章于 2019-08-21 21:05:00 发布

just_sort

最新推荐文章于 2019-08-21 21:05:00 发布

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本文链接：https://blog.youkuaiyun.com/just_sort/article/details/52414460

本文介绍了一种通过矩形切割的方法来解决矩形面积并的问题，即计算多个矩形在平面上叠加后的总面积。该问题常见于算法竞赛中，解决思路清晰且附带AC代码。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18556		Accepted: 5115

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.

【题意】就是矩形面积并。

【解题方法】矩形切割。

【AC 代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 40005;
struct node{
    LL x1,y1;
    LL x2,y2;
    LL sum;
//    void read()
//    {
//        scanf("%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2);
//        sum=0LL;
//    }
}T[maxn];
LL n;
bool cmp(node aa,node bb)
{
    if(aa.y2!=bb.y2) return aa.y2<bb.y2;
    else{
        if(aa.x2!=bb.x2) return aa.x2<bb.x2;
        else             return aa.x1>bb.x1;
    }
}
void Cover(LL x1,LL y1,LL x2,LL y2,LL k,LL c)
{
    while(k<n&&(x1>=T[k].x2||x2<=T[k].x1||y1>=T[k].y2||y2<=T[k].y1)) k++;
    if(k>=n){
        T[c].sum+=(x2-x1)*(y2-y1);
        return ;
    }
    if(x1<T[k].x1){
        Cover(x1,y1,T[k].x1,y2,k+1,c);
        x1=T[k].x1;
    }
    if(x2>T[k].x2){
        Cover(T[k].x2,y1,x2,y2,k+1,c);
        x2=T[k].x2;
    }
    if(y1<T[k].y1){
        Cover(x1,y1,x2,T[k].y1,k+1,c);
        y1=T[k].y1;
    }
    if(y2>T[k].y2){
        Cover(x1,T[k].y2,x2,y2,k+1,c);
        y2=T[k].y2;
    }
}
int main()
{
    LL a,b,c;
    while(scanf("%I64d",&n)!=EOF)
    {
        for(int i=0; i<n; i++){
            scanf("%I64d%I64d%I64d",&a,&b,&c);
            T[i].x1=a,T[i].y1=0LL;
            T[i].x2=b,T[i].y2=c;
            T[i].sum=0;
        }
        sort(T,T+n,cmp);
        //
        //system("pause");
        for(LL i=n-1; i>=0; i--){
            Cover(T[i].x1,T[i].y1,T[i].x2,T[i].y2,i+1,i);
        }
        LL ans=0;
        for(int i=0; i<n; i++){
            ans+=T[i].sum;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}