zoj 1128 || poj 1151 Atlantis

求矩形并的面积。



传说中的离散化。



因为矩形最多才100个，所以将所有的x坐标从小到大映射下，y也是。数组map[I][K]相当于横坐标为x[I] 纵坐标为y[K]的点，也就相当于把很大距离的点缩到了200*200上，然后再扫描下求面积即可。



引用下PPT的这个图。。

#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX = 205;
const double eps = 1e-6;
struct rectangle{ double lx,ly,rx,ry;};
rectangle r[MAX];
double x[MAX],y[MAX];
int map[MAX][MAX];
bool dy(double x,double y)	{	return x > y + eps;}	// x > y 
bool xy(double x,double y)	{	return x < y - eps;}	// x < y 
bool dyd(double x,double y)	{ 	return x > y - eps;}	// x >= y 
bool xyd(double x,double y)	{	return x < y + eps;} 	// x <= y 
bool dd(double x,double y) 	{	return fabs( x - y ) < eps;}  // x == y
int find(double a[],double x,int m)
{
	int beg = 0,end = m-1;
	while( beg <= end )
	{
		int mid = (beg+end) >> 1;
		if( dd(x,a[mid]) )
			return mid;
		if( dy(x,a[mid]) )
			beg = mid + 1;
		else
			end = mid;
	}
}
double area_rectangle(int m,int n)
{
	for(int i=0; i<n; i++)
	{
		int x1 = find(x,r[i].lx,m);
		int y1 = find(y,r[i].ly,m);
		int x2 = find(x,r[i].rx,m);
		int y2 = find(y,r[i].ry,m);
		for(int k=x1; k<x2; k++)
			for(int j=y1; j<y2; j++)
				map[k][j] = true;
	}
	double sum = 0.0;
	for(int i=0; i<m-1; i++)
		for(int k=0; k<m-1; k++)
			if( map[i][k] )
				sum += (x[i+1] - x[i])*(y[k+1] - y[k]);
	return sum;
}
int main()
{
	int n;
	int ind = 1;
	while( ~scanf("%d",&n) && n )
	{
		int k = 0;
		memset(map,false,sizeof(map));
		for(int i=0; i<n; i++)
		{
			scanf("%lf%lf%lf%lf",&r[i].lx,&r[i].ly,&r[i].rx,&r[i].ry);
			x[k] = r[i].lx;
			y[k++] = r[i].ly;
			x[k] = r[i].rx;
			y[k++] = r[i].ry;
		}
		sort(x,x+k);
		sort(y,y+k);
		double area = area_rectangle(k,n);
		printf("Test case #%d/nTotal explored area: %.2lf/n/n",ind++,area);
	}
return 0;
}