HDU 1572 Atlantis 線段樹掃描線（矩形面積合併）

HDU 1572 Atlantis

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

Input

The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don’t process it.

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.

Sample Input

2
10 10 20 20
15 15 25 25.5
0

Sample Output

Test case #1
Total explored area: 180.00 

題意：

給定 n 個矩形的對角線座標，求他們所覆蓋的面積（會有重疊）

矩形面積合併

線段樹掃描線：（首先將 x端點離散化，方便計算 ）將一個矩形拆成兩條線（左右邊界的橫座標爲端點，兩個縱座標定位兩條線，上線標記-1，下線標記1），這些線按照 縱座標y 從小到大排序，然後去按順序遍歷每一條線，更新線段樹：線段樹記錄的是tree.l 到 tree.r區間覆蓋了多少次+1-1，將它們疊加到一起，tree,sum 記錄這個區間的長度

看圖吧，一眼明瞭

擼代碼：

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct line
{
    double x1,x2,y,l;
}p[55220];
struct Tree
{
    int l,r,ans;
    double sum;
}tree[51010];

bool comp(line a,line b)
{
    return a.y<b.y;
}
double x[55220];
void build(int l,int r,int root)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].ans=0;
    tree[root].sum=0;
    if(l==r)
        return ;
    int mid = (l+r)>>1;
    build(l,mid,root<<1);
    build(mid+1,r,root<<1|1);
    return ;
}

void pushdown(int root)
{
    if(tree[root].ans!=0) tree[root].sum=x[tree[root].r+1]-x[tree[root].l];/**計算長度，而不是點*/
    else if(tree[root].l==tree[root].r) tree[root].sum=0;/**葉子結點,且覆蓋0次*/
    else tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;/**非葉子結點，ans=0*/
    return ;
}
void update(int l,int r,int kk,int root)
{
    if(l<=tree[root].l&&r>=tree[root].r)
    {
        tree[root].ans+=kk;/**tree.l 到 tree.r區間覆蓋了+1或-1*/
        pushdown(root);/**更新sum值*/
        return ;
    }
    int mid=(tree[root].l+tree[root].r)>>1;
    if(l<=mid)update(l,r,kk,root<<1);
    if(mid<r)update(l,r,kk,root<<1|1);
    pushdown(root);
    return ;
}
int main()
{
    int n,t=1;
    while(~scanf("%d",&n)&&n)
    {
        double x1,y1,x2,y2;
        double ans=0;
        int m,k=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            p[k].x1=x1,p[k].x2=x2;
            p[k].y=y1;
            p[k].l=1;
            x[k++]=x1;

            p[k].x1=x1,p[k].x2=x2;
            p[k].y=y2;
            p[k].l=-1;
            x[k++]=x2;
        }
        sort(x,x+k);
        sort(p,p+k,comp);
        m=1;
        for(int i=1;i<k;i++)
        {
            if(x[i]!=x[i-1])
                x[m++]=x[i];
        }
        build(0,m-2,1);/*初始化，m個點 ，就有m-1個線段？*/
        ans=0;
        for(int i=0;i<k-1;i++)
        {
            int l=lower_bound(x,x+m,p[i].x1)-x;
            int r=lower_bound(x,x+m,p[i].x2)-x-1;/**離散化處理,線段而不是點*/
            if(l<=r)/*!!!不用應該也行*/
                update(l,r,p[i].l,1);
            ans+=tree[1].sum*(p[i+1].y-p[i].y);/*所求線的不重合的寬度*/
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n",t++,ans);

    }
    return 0;
}