HDU 1542 Atlantis 扫描线-优快云博客

本文介绍了一种通过构建线段树来计算多个矩形并集面积的方法。使用C++实现，具体步骤包括：对输入的矩形边界进行排序、构建线段树以维护区间覆盖状态、更新线段树反映矩形的增加或移除，并计算最终的并集面积。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1542
题意：裸的求长方形的并

代码：

#include <bits/stdc++.h>
#define sf scanf
#define pf printf

using namespace std;
const int maxn = 100 + 5;
const int TOT_SEG = maxn * 2;
struct Seg{
    int f;
    double l,r,h;
    void set(double ll,double rr,double hh,int d){
        l = ll,r = rr,h = hh,f = d;
    }
}Segs[TOT_SEG];

bool cmp(const Seg& a,const Seg& b){
    return a.h < b.h;
}

double AR[TOT_SEG]; /** 区间长度 */
int FALG[TOT_SEG];
/** SEG_TREE */
#define lson rt << 1 , l , mid
#define rson rt << 1 | 1,mid + 1,r
double _SUM[TOT_SEG << 2];
void PushUp(int rt){
    _SUM[rt] = _SUM[rt << 1] + _SUM[rt << 1 | 1];
}
void Build(int rt,int l,int r){
    if(l == r){
        _SUM[rt] = FALG[l] = 0;
        return;
    }
    int mid = l + r >> 1;
    Build(lson),Build(rson);
}

void Update(int rt,int l,int r,int L,int R,int d){
    if(l == r){
        FALG[l] += d;
        if(FALG[l] == 0) _SUM[rt] = 0;
        else _SUM[rt] = AR[r];
        return;
    }
    int mid = l + r >> 1;
    if(L <= mid) Update(lson,L,R,d);
    if(R > mid) Update(rson,L,R,d);
    PushUp(rt);
}

map<double,int> POS;
int main(){
    int n,ca = 0;
    while( ~sf("%d",&n) && n ){
        int tmp = 1;double x1,x2,y1,y2;
        for(int i = 0;i < n;++i){
            sf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
            Segs[tmp].set(x1,x2,y1,1);
            AR[tmp++] = x1;
            Segs[tmp].set(x1,x2,y2,-1);
            AR[tmp++] = x2;
        }
        POS.clear();
        sort(Segs + 1,Segs + tmp,cmp);
        sort(AR + 1,AR + tmp);
        AR[tmp] = -1;
        int AR_TMP = 2;
        for(int i = 2;i < tmp;++i)
            if(AR[i] != AR[i - 1]){
                AR[AR_TMP] = AR[i];
                POS[AR[i]] = AR_TMP++;
            }
        for(int i = 1;i < AR_TMP;++i){
            AR[i] = AR[i + 1] - AR[i];
        }
        Build(1,1,AR_TMP - 2);
        double ans = 0;
        for(int i = 1;i < tmp;++i){
            int L = POS[Segs[i].l],R = POS[Segs[i].r] - 1;
            Update(1,1,AR_TMP - 2,L,R,Segs[i].f);
            ans = ans + _SUM[1] * (Segs[i + 1].h - Segs[i].h);
        }
        pf("Test case #%d\n",++ca);
        pf("Total explored area: %.2lf\n\n",ans);
    }

}