wikioi p3044 矩形面积求并

同样的离散化，题解wiki上有。

略。

#include<stdio.h>
#include<iostream>
#include<memory.h>
#include<algorithm>
using namespace std;
const int MAX_N = 201;
int N;
bool f[MAX_N][MAX_N];
double x[MAX_N];
double y[MAX_N];
int lengthx;
int lengthy;
double ans;
struct node
{
	float x1,x2,y1,y2;
}G[MAX_N];
int init()
{
	memset(x,0,sizeof(x));
	memset(y,0,sizeof(y));
	memset(f,false,sizeof(f));
	lengthx=lengthy=0;
    ans=0;
	int i;
	for (i=1;i<=N;i++)
	{
		scanf("%f %f %f %f",&G[i].x1,&G[i].y1,&G[i].x2,&G[i].y2);
		x[i*2-1]=G[i].x1;
		x[i*2]=G[i].x2;
		y[i*2-1]=G[i].y1;
		y[i*2]=G[i].y2;
	}
	sort(x+1,x+2*N+1);
	sort(y+1,y+2*N+1);
	
	int j,k,a,b;
	for (i=1;i<=N;i++)
	{
		a=1;
		b=1;
		while (x[a]<=G[i].x1&&a<=2*N) a++;
		while (y[b]<=G[i].y1&&b<=2*N) b++;
		for (j=a;x[j]<=G[i].x2&&j<=2*N;j++)
		for (k=b;y[k]<=G[i].y2&&k<=2*N;k++)
		f[j][k]=true;
	}
	for (i=1;i<=N*2;i++)
	for (j=1;j<=N*2;j++)
	if (f[i][j]) ans+=(x[i]-x[i-1])*(y[j]-y[j-1]);
}
int put()
{
	printf("%.2f\n",ans);
}
int work()
{
	while (1)
	{
        scanf("%d",&N);
        if (!N) return 0;
		init();
		put();
	}
}
int main()
{
	work();
	return 0;
}
#include<stdio.h>
#incl