HDU1255 覆盖的面积 （扫描线）

覆盖的面积

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5117 Accepted Submission(s): 2568

Problem Description
给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.

Input
输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐标的范围从0到100000.

注意:本题的输入数据较多,推荐使用scanf读入数据.

Output
对于每组测试数据,请计算出被这些矩形覆盖过至少两次的区域的面积.结果保留两位小数.

Sample Input
2
5
1 1 4 2
1 3 3 7
2 1.5 5 4.5
3.5 1.25 7.5 4
6 3 10 7
3
0 0 1 1
1 0 2 1
2 0 3 1

Sample Output
7.63
0.00

Author
Ignatius.L & weigang Lee

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基础扫描线

#include 
#include 
#include 
#include 
#include 
#include 
#define MAX 2005
using namespace std;
int cntline=0;
struct type_line
{
    double left,right,y;
    int mark;
}line[MAX];
bool cmp(const type_line &a,const type_line &b)
{
    return a.yvector <double> list;
struct node
{
    int left,right,mark;
    double sum1,sum2;
}tree[MAX*4];

void addline(double left,double right,double y,int mark)
{
    line[cntline].left=left;
    line[cntline].right=right;
    line[cntline].y=y;
    line[cntline].mark=mark;
    cntline++;
}
void build(int left,int right,int root)
{
    tree[root].left=left;
    tree[root].right=right;
    tree[root].sum1=tree[root].sum2=0;
    tree[root].mark=0;
    if(left+1==right)
        return;
    else
    {
        int mid=(left+right)/2;
        build(left,mid,root*2);
        build(mid,right,root*2+1);
    }
}
void update(int left,int right,int root,int mark)
{
    int l=tree[root].left;
    int r=tree[root].right;
    if(l>=left&&r<=right)
    {
        tree[root].mark+=mark;
    }
    else
    {
        int mid=(tree[root].left+tree[root].right)/2;
        if(left2,mark);
        if(right>mid)
            update(left,right,root*2+1,mark);
    }
    if(tree[root].mark)
    {
        tree[root].sum1=list[r]-list[l];
    }
    else if(l+1==r)
        tree[root].sum1=0;
    else
        tree[root].sum1=tree[root*2].sum1+tree[root*2+1].sum1;

    if(tree[root].mark>1)
        tree[root].sum2=list[r]-list[l];
    else if(l+1==r)
        tree[root].sum2=0;
    else if(tree[root].mark==1)
        tree[root].sum2=tree[root*2].sum1+tree[root*2+1].sum1;
    else
        tree[root].sum2=tree[root*2].sum2+tree[root*2+1].sum2;

}

int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        cntline=0;
        list.clear();
        double x1,y1,x2,y2;
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            if(x1==x2)
                continue;
            addline(x1,x2,y1,1);
            addline(x1,x2,y2,-1);
            list.push_back(x1);
            list.push_back(x2);
        }
        sort(line,line+cntline,cmp);
        sort(list.begin(),list.end());
        list.erase(unique(list.begin(),list.end()),list.end());
        build(0,list.size()-1,1);
        double ans=0;
        for(int i=0;i<=cntline-1;i++)
        {
            int left=lower_bound(list.begin(),list.end(),line[i].left)-list.begin();
            int right=lower_bound(list.begin(),list.end(),line[i].right)-list.begin();
            if(i>0)
                ans+=tree[1].sum2*(line[i].y-line[i-1].y);
            update(left,right,1,line[i].mark);
        }
        printf("%.2lf\n",ans);

    }
}