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最新推荐文章于 2023-01-02 16:19:02 发布

原创最新推荐文章于 2023-01-02 16:19:02 发布 · 464 阅读

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文章标签：

#tree #build #fun #up #struct #im

数据结构同时被 2 个专栏收录

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线段树

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本文介绍了一种使用线段树的数据结构来解决二维空间中多个矩形覆盖区域的面积计算问题。通过扫描线算法，结合线段树更新和查询，实现了高效求解。适用于计算机图形学及算法竞赛等领域。

面积并：


#include <iostream>
#include <string.h>
using namespace std;
struct node
{
  int l;
  int r;
  int cover;
  double once;
  double len;
};
node tree[7000];
struct Line
{
   double x;
   double y_down;
   double y_up;
   int cover;
};
Line line[7000];
double yy[7000];
bool cmp(Line a,Line b)
{
    if( a.x == b.x ) 
       return a.cover > b.cover;
    else
       return a.x < b.x;
}
void Build(int i,int l,int r)
{
    tree[i].l = l; 
    tree[i].r = r;
    tree[i].cover = 0;
    tree[i].once = 0.0;
    if(tree[i].l == tree[i].r - 1 )
        return ;
    int mid =(l+r)/2;
    Build(i*2,l,mid);
    Build(i*2+1,mid,r);
}
void fun(int i)
{
    if( tree[i].cover >= 2 )
    {
        tree[i].once = 0;
        tree[i].len = yy[tree[i].r] - yy[tree[i].l];
        return ;
    }
    if( tree[i].cover == 1 )
    {
        if( tree[i].l == tree[i].r - 1 )
            tree[i].len = 0;
        else
            tree[i].len = tree[2*i+1].len + tree[2*i].len + tree[2*i+1].once + tree[2*i].once;
        tree[i].once = yy[tree[i].r] - yy[tree[i].l] - tree[i].len;
        return ;    
    }
    if( tree[i].cover == 0 )
    {
        if( tree[i].l == tree[i].r - 1 )
            tree[i].len =tree[i].once = 0;
        else
        {
            tree[i].len = tree[2*i+1].len + tree[2*i].len;
            tree[i].once =tree[2*i+1].once+tree[2*i].once;
        }
        return ;
    }
}
void Updata(int i, Line p)
{
    if( yy[tree[i].l] >= p.y_down && yy[tree[i].r] <= p.y_up )
    {
        tree[i].cover += p.cover;
        fun(i);
        return ;
    }
    if( tree[i].l == tree[i].r - 1 ) return ;
    int mid = (tree[i].l+tree[i].r)/2;
    if( p.y_down <= yy[mid] )
        Updata(2*i,p);
    if( p.y_up > yy[mid] )
        Updata(2*i+1,p);
    fun(i);
}
int main()
{
    int t,n;
    double x1,x2,y1,y2;
    scanf("%d",&t);
    while( t-- )
    {
        scanf("%d",&n);
        int m = 0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[m].x = x1; 
            line[m].y_down = y1; 
            line[m].y_up = y2;
            line[m].cover = 1;
             yy[m++] = y1;
            line[m].x = x2; 
            line[m].y_down = y1; 
            line[m].y_up = y2;
            line[m].cover = -1;
            yy[m++] = y2;
        }
        sort(yy,yy+m);
        sort(line,line+m,cmp);
        int i = unique(yy,yy+m) - yy;
        Build(1,0,i-1);
        double ans = 0;
        Updata(1,line[0]);
        for(int i=1; i<m; i++)
        {
           ans += tree[1].len*(line[i].x - line[i-1].x);
           Updata(1,line[i]);
        }
        printf("%.2lf\n",ans);
    }

return 0;
}