poj 1151 Atlantis 离散化

/*



题目：

    给出n个矩形的坐标，求所有的矩形的覆盖面积



分析：

    离散化,具体请看Matrix67  http://www.matrix67.com/blog/archives/108



*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>



using namespace std;



const int X = 205;



double x[X],y[X];

int topx,topy;

int n;

double lx[X],rx[X],ly[X],ry[X];

bool map[X][X];



int posx(double a)

{

    for(int i=0;i<topx;i++)

        if(x[i]==a)

            return i;

    return topx;

}



int posy(double a)

{

    for(int i=0;i<topy;i++)

        if(y[i]==a)

            return i;

    return topy;

}



void addx(double a)

{

    int temp = posx(a);

    if(temp==topx)

        x[topx++] = a;

}



void addy(double a)

{

    int temp = posy(a);

    if(temp==topy)

        y[topy++] = a;

}



int main()

{

    freopen("sum.in","r",stdin);

    int cnt = 0;

    while(cin>>n,n)

    {

        topx = topy = 0;

        for(int i=0;i<n;i++)

        {

            cin>>lx[i]>>ly[i]>>rx[i]>>ry[i];

            addx(lx[i]);

            addx(rx[i]);

            addy(ly[i]);

            addy(ry[i]);

        }

        sort(x,x+topx);

        sort(y,y+topy);

        double ans = 0;

        memset(map,false,sizeof(map));

        int x1,x2,y1,y2;

        for(int i=0;i<n;i++)

        {

            x1 = posx(lx[i]);

            x2 = posx(rx[i]);

            y1 = posy(ly[i]);

            y2 = posy(ry[i]);

            for(int j=x1+1;j<=x2;j++)

                for(int k=y1+1;k<=y2;k++)

                    map[j][k] = true;

        }

        for(int i=0;i<topx;i++)

            for(int j=0;j<topy;j++)

                if(map[i][j])

                    ans += (x[i]-x[i-1])*(y[j]-y[j-1]);

        printf("Test case #%d\n",++cnt);

        printf("Total explored area: %.2lf\n\n",ans);

    }

    return 0;

}