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Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of
all rectangles is called the perimeter.

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.

The corresponding boundary is the whole set of line segments drawn in Figure 2.

The vertices of all rectangles have integer coordinates.

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The
values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.

0 <= number of rectangles < 5000

All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Please process to the end of file.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

228

Source

IOI 1998

轮廓线确实有些屌………………………………

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int hs[10010];
struct Tree
{
	int l,r,num,cnt,val;
	bool flag_l,flag_r;
}tree[10000<<2];
void create(int l,int r,int k)
{
	tree[k].l=l;
	tree[k].r=r;
	tree[k].val=0;
	tree[k].num=tree[k].cnt=0;
	tree[k].flag_l=tree[k].flag_r=0;
	if(l+1==r)
		return;
	int m=l+r>>1;
	create(l,m,k<<1);
	create(m,r,k<<1|1);
}
void update(int l,int r,int k,int flag)
{
	if(tree[k].l==l&&tree[k].r==r)
	{
		tree[k].num+=flag;
		if(tree[k].num==0)
		{
			tree[k].val=l+1==r?0:tree[k<<1].val+tree[k<<1|1].val;
			if(l+1==r)
			{
				tree[k].cnt=0;
				tree[k].flag_l=tree[k].flag_r=0;
			}
			else
			{
				tree[k].cnt=tree[k<<1].cnt+tree[k<<1|1].cnt;
				if(tree[k<<1].flag_r&&tree[k<<1|1].flag_l)
					tree[k].cnt--;
				tree[k].flag_l=tree[k<<1].flag_l;
				tree[k].flag_r=tree[k<<1|1].flag_r;
			}
		}
		else
		{
			tree[k].cnt=1;
			tree[k].val=hs[r]-hs[l];
			tree[k].flag_l=tree[k].flag_r=1;
		}
		return;
	}
	int m=tree[k].l+tree[k].r>>1;
	if(r<=m)
		update(l,r,k<<1,flag);
	else if(l>=m)
		update(l,r,k<<1|1,flag);
	else
	{
		update(l,m,k<<1,flag);
		update(m,r,k<<1|1,flag);
	}
	if(tree[k].num==0)
	{
		tree[k].val=tree[k<<1].val+tree[k<<1|1].val;
		tree[k].cnt=tree[k<<1].cnt+tree[k<<1|1].cnt;
		if(tree[k<<1].flag_r&&tree[k<<1|1].flag_l)
			tree[k].cnt--;
		tree[k].flag_l=tree[k<<1].flag_l;
		tree[k].flag_r=tree[k<<1|1].flag_r;
	}
}
struct Seg
{
	int flag,l,r,y;
	bool operator <(Seg one)const
	{
		return y<one.y;
	}
}seg[10010];
int main()
{
	int n;
	while(cin>>n)
	{
		vector<int>box;
		for(int i=0;i<n;i++)
		{
			int x1,y1,x2,y2;
			cin>>x1>>y1>>x2>>y2;
			seg[i<<1].l=seg[i<<1|1].l=x1;
			seg[i<<1].r=seg[i<<1|1].r=x2;
			seg[i<<1].y=y1;
			seg[i<<1|1].y=y2;
			seg[i<<1].flag=1;
			seg[i<<1|1].flag=-1;
			box.push_back(x1);
			box.push_back(x2);
		}
		sort(box.begin(),box.end());
		box.erase(unique(box.begin(),box.end()),box.end());
		for(int i=0;i<box.size();i++)
			hs[i+1]=box[i];
		n*=2;
		sort(seg,seg+n);
		int ans=0,p=0;
		create(1,box.size(),1);
		for(int i=0;i<n;i++)
		{
			int l=1+lower_bound(box.begin(),box.end(),seg[i].l)-box.begin();
			int r=1+lower_bound(box.begin(),box.end(),seg[i].r)-box.begin();
			update(l,r,1,seg[i].flag);
			ans+=abs(tree[1].val-p);
			p=tree[1].val;
			if(i!=n-1)
				ans+=(seg[i+1].y-seg[i].y)*tree[1].cnt*2;
		}
		printf("%d\n",ans);
	}
}