Atlantis
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 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
Source：
Mid-Central European Regional Contest 2000
题意：
给你n个矩形，每个矩形给出左下点的坐标，右上点的坐标。最后以n=0为结束。要你求出矩形并后的面积。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=210;
int n,k,cas,xy[maxn][maxn];
double sum,x[maxn],y[maxn],s[maxn][5];
int main()
{
while(cin>>n)
{
if(!n) break;
cas++;
k=1,sum=0.0;
memset(xy,0,sizeof(xy));
for(int i=1;i<=n;i++)
{
cin>>s[i][1]>>s[i][2]>>s[i][3]>>s[i][4];
x[k]=s[i][1];
y[k]=s[i][2];
k++;
x[k]=s[i][3];
y[k]=s[i][4];
k++;
}
sort(x+1,x+n*2+1);
sort(y+1,y+n*2+1);
for(int k=1;k<=n;k++)
{
int i1,i2,j1,j2;
for(i1=1;i1<=2*n;i1++)
if(x[i1]==s[k][1]) break;
for(i2=1;i2<=2*n;i2++)
if(x[i2]==s[k][3]) break;
for(j1=1;j1<=2*n;j1++)
if(y[j1]==s[k][2]) break;
for(j2=1;j2<=2*n;j2++)
if(y[j2]==s[k][4]) break;
for(int i=i1;i<i2;i++)
for(int j=j1;j<j2;j++)
xy[i][j]=1;
}
for(int i=1;i<=n*2;i++)
for(int j=1;j<=n*2;j++)
sum+=xy[i][j]*(x[i+1]-x[i])*(y[j+1]-y[j]);
printf("Test case #%d
",cas);
printf("Total explored area: %.2f
",sum);
printf("
");
}
return 0;
}