hdu1542 && pku 1151 Atlantis（面积并）

求N个矩形的面积并，这个比周长并要简单的多，不过涉及到了离散化，根据相对大小，给对应的double型数据编号，插入是再二分查找编号即可

也可以用矩形切割去做，代码简单很多

#include<iostream>
#include<algorithm>
#include<string>
#include<math.h>
using namespace std;
const int N = 100+10;
struct rec
{
    double p1[2],p2[2];
}r[N];
rec rr[N*N];
int total;
inline double get_area(rec& a)
{
    return (a.p2[0]-a.p1[0])*(a.p2[1]-a.p1[1]);
}
inline bool cut_rec(rec &now,rec t)//矩形切割
{
     for(int i=0;i<2;i++)
         if(t.p1[i]>=now.p2[i] || t.p2[i]<=now.p1[i])
             return false;
     rec tmp;
     double k1,k2;
     for(int i=0;i<2;i++)
     {
         k1=max(t.p1[i],now.p1[i]);
         k2=min(t.p2[i],now.p2[i]);
         if(t.p1[i]<k1)
         {
             tmp=t;
             tmp.p2[i]=k1;
             rr[total++]=tmp;
         }
         if(t.p2[i]>k2)
         {
             tmp=t;
             tmp.p1[i]=k2;
             rr[total++]=tmp;
         }
         t.p1[i]=k1;
         t.p2[i]=k2;
     }
     return true;
}
int main()
{
    int n,cas=0;
    while(scanf("%d",&n)==1 && n)
    {
        for(int i=0;i<n;i++)
            scanf("%lf %lf %lf %lf",&r[i].p1[0],&r[i].p1[1],&r[i].p2[0],&r[i].p2[1]);
        total=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<total;j++)
            {
                if(cut_rec(r[i],rr[j]))
                {
                    rr[j]=rr[total-1];
                    total--;
                    j--;
                }
            }
            rr[total++]=r[i];
        }
        double ans=0;

        for(int i=0;i<total;i++)
            ans+=get_area(rr[i]);
        printf("Test case #%d\n",++cas);
        printf("Total explored area: %.2f\n\n",ans);
    }
    return 0;
}

#include<iostream>
#include<algorithm>
#include<map>
#define maxn 222
using namespace std;
struct node
{
	double x,y1,y2;
	int s;
	node(double a=0,double b=0,double c=0,int d=0):x(a),y1(b),y2(c),s(d){}
	friend bool operator<(const node a,const node b)
	{
		return a.x<b.x;
	}
};
node ss[maxn];
bool cmp(node a,node b)
{
	return a.x<b.x;
}
double len[maxn<<2];
int cnt[maxn<<2];
double map1[maxn];
void PushUp(int k,int s,int t)
{
	if(cnt[k])
		len[k]=map1[t+1]-map1[s];
	else if(t==s)
		len[k]=0;
	else len[k]=len[k<<1]+len[k<<1 |1];
}
void update(int l,int r,int c,int s,int t,int k)
{
	if(l<=s && t<=r)
	{
		cnt[k]+=c;
		PushUp(k,s,t);
		return ;
	}
	int kl=k<<1,kr=kl+1,mid=(s+t)>>1;
	if(l<=mid)
		update(l,r,c,s,mid,kl);
	if(r>mid)
		update(l,r,c,mid+1,t,kr);
	PushUp(k,s,t);
}
int Bin(double key,int n,double map1[]) {
	int l = 0 , r = n - 1;
	while (l <= r) 
	{
		int mid = (l + r) >> 1;
		if (map1[mid] == key) return mid;
		if (map1[mid] < key) l = mid + 1;
		else r = mid - 1;
	}
	return -1;
}
int main()
{
	double a,b,c,d;
	int n,cas=0;
	while(scanf("%d",&n)==1 &&n)
	{
		int m=0;
		for(int i=0;i<n;i++)
		{
			scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
			map1[m]=b;
			ss[m++]=node(a,b,d,1);
	    	map1[m]=d;
			ss[m++]=node(c,b,d,-1);
		}
		sort(map1,map1+m);
		sort(ss,ss+m);
		int k=1;
		for (int i = 1 ; i < m ; i ++) 
			if (map1[i] != map1[i-1])
				map1[k++] =map1[i];
		memset(cnt , 0 , sizeof(cnt));
		memset(len , 0 , sizeof(len));
		double ans=0;
		for(int i=0;i<m-1;i++)
		{
			int l = Bin(ss[i].y1 , k , map1);
			int r = Bin(ss[i].y2 , k , map1) - 1;
			if (l <= r)
				update(l , r , ss[i].s , 0 , k - 1, 1);
			ans+=len[1]*(ss[i+1].x-ss[i].x);
		}
		printf("Test case #%d\nTotal explored area: %.2lf\n\n",++cas,ans);
	}
	return 0;
}