Atlantis

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/39561889

本文介绍了一个计算多个矩形区域合并后总面积的问题，并提供了一种使用扫描线算法的解决方案。通过输入多个矩形的地图坐标，程序能够计算出这些矩形覆盖的总面积。

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7210 Accepted Submission(s): 3173

Problem Description

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 file 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

题意：给你n个矩阵，问矩阵合并之后的总面积。

思路：裸扫描线。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double q[21000];
struct node
{
    int l,r,val;
    double len;
}tree[81000];
struct node2
{
    double x1,x2,h;
    int d;
}op[21000];
bool cmp(node2 a,node2 b)
{
    return a.h<b.h;
}
int n,m;
void build(int o,int l,int r)
{
    tree[o].l=l;
    tree[o].r=r;
    tree[o].val=0;
    tree[o].len=0;
    if(l==r)
      return;
    int mi=(l+r)/2;
    build(o*2,l,mi);
    build(o*2+1,mi+1,r);
}
void up(int o)
{
    if(tree[o].val>0)
      tree[o].len=q[tree[o].r]-q[tree[o].l-1];
    else if(tree[o].l==tree[o].r)
      tree[o].len=0;
    else
      tree[o].len=tree[o*2].len+tree[o*2+1].len;
}
void update(int o,int l,int r,int val)
{
    if(tree[o].l==l && tree[o].r==r)
    {
        tree[o].val+=val;
        up(o);
        return;
    }
    int mi=(tree[o].l+tree[o].r)/2;
    if(r<=mi)
      update(o*2,l,r,val);
    else if(l>mi)
      update(o*2+1,l,r,val);
    else
    {
        update(o*2,l,mi,val);
        update(o*2+1,mi+1,r,val);
    }
    up(o);
}
int find(double a)
{
    int l=1,r=m,mi;
    while(l<r)
    {
        mi=(l+r)/2;
        if(q[mi]<a)
          l=mi+1;
        else
          r=mi;
    }
    return l;
}
int main()
{
    int t=0,i,j,k;
    double x1,x2,y1,y2,ans;
    while(~scanf("%d",&n) && n)
    {
        t++;
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            op[i*2-1].x1=x1;op[i*2-1].x2=x2;op[i*2-1].d=-1;op[i*2-1].h=y2;
            op[i*2].x1=x1;op[i*2].x2=x2;op[i*2].d=1;op[i*2].h=y1;
            q[i*2-1]=x1;q[i*2]=x2;
        }
        sort(q+1,q+1+n*2);
        sort(op+1,op+1+n*2,cmp);
        op[0].h=0;
        m=1;
        for(i=2;i<=n*2;i++)
           if(q[i]!=q[i-1])
             q[++m]=q[i];
        build(1,1,m);
        ans=0;
        for(i=1;i<n*2;i++)
        {
            update(1,find(op[i].x1)+1,find(op[i].x2),op[i].d);
            ans+=tree[1].len*(op[i+1].h-op[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n",t,ans);
    }
}