线段树+扫描线+离散化解poj1151 hdu 1542 ( Atlantis )

受此链接很大启发才明白扫描线：

http://www.cnblogs.com/scau20110726/archive/2013/04/12/3016765.html

我的代码如下：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<double,double> ll;
vector<double>x;
int n;
struct node
{
	double l,r,y;
	int flag;
}seg[201];
struct node1
{
	int l,r,m;
	double y;
	int flag;
}tree[801];

bool cmp(double a,double b)
{
	return a<b;
}
bool cmp1(node a,node b)
{
	return a.y<b.y;
}

void create(int l,int r,int k)
{
	tree[k].l=l;
	tree[k].r=r;
	tree[k].m=(l+r)>>1;
	tree[k].flag=0;
	if(l+1==r)
		return;
	create(l,tree[k].m,k<<1);
	create(tree[k].m,r,k<<1|1);
}
void join(int l,int r,int flag,int k)
{
	if(tree[k].l==l&&tree[k].r==r)
	{
		tree[k].flag+=flag;
		return;
	}
	if(r<=tree[k].m)
		join(l,r,flag,k<<1);
	else if(l>=tree[k].m)
		join(l,r,flag,k<<1|1);
	else
	{
		join(l,tree[k].m,flag,k<<1);
		join(tree[k].m,r,flag,k<<1|1);
	}
}
double find(int k)
{
	if(tree[k].flag>0)
		return x[tree[k].r]-x[tree[k].l];
	if(tree[k].l+1==tree[k].r)
		return 0;
	return find(k<<1)+find(k<<1|1);
}
void in()
{
	ll t1,t2;
	int i=0;
	x.clear();
	while(n--)
	{
		cin>>t1.first>>t1.second>>t2.first>>t2.second;
		seg[i].l=t1.first;
		seg[i].r=t2.first;
		seg[i].y=t1.second;
		seg[i].flag=1;
		x.push_back(t1.first);
		i++;
		seg[i].l=t1.first;
		seg[i].r=t2.first;
		seg[i].y=t2.second;
		seg[i].flag=-1;
		x.push_back(t2.first);
		i++;
	}
	n=i;
	sort(x.begin(),x.end(),cmp);
	x.erase(unique(x.begin(),x.end()),x.end());
	create(0,n-1,1);
}
double work()
{
	double ans=0,t1;
	int i,l,r;
	sort(seg,seg+n,cmp1);
	n--;
	for(i=0;i<n;i++)
	{
		l=lower_bound(x.begin(),x.end(),seg[i].l)-x.begin();
		r=lower_bound(x.begin(),x.end(),seg[i].r)-x.begin();
		join(l,r,seg[i].flag,1);
		t1=find(1);
		ans+=find(1)*(seg[i+1].y-seg[i].y);
	}
	return ans;
}
int main()
{
	int jishu=0;
	while(cin>>n&&n!=0)
	{
		in();
		printf("Test case #%d\nTotal explored area: %.2lf\n\n",++jishu,work());
	}
}

线段树+扫描线+离散化解poj1151 hdu 1542 ( Atlantis ),布布扣,bubuko.com

线段树+扫描线+离散化解poj1151 hdu 1542 ( Atlantis )

原文：http://blog.csdn.net/stl112514/article/details/38585673