HDU1828--Picture

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最新推荐文章于 2025-05-16 10:00:00 发布

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分类专栏：数据结构_线段树文章标签： HDU1828 Picture

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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

/*

请无视这段代码。直接看后面优化好的代码~~~

*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
#define maxn 10008
int X[maxn],XX[maxn];
int Y[maxn],YY[maxn];//一个用来存原始的，一个用来存离散后的
//矩形的个数5000上限，横坐标上限个数10000个
map <int , int> collx;
map <int , int> colly;
int n;
int tree;
void lisanx(int * x)
{
	sort(x+1,x+2*n+1);
	tree=0;
	for(int i=1;i<=2*n;i++)
	{
		if(collx[x[i]]==0)
		{
			collx[x[i]]=++tree;
			XX[tree]=x[i];
		}
	}
}
void lisany(int * y)
{
	sort(y+1,y+2*n+1);
	tree=0;
	for(int i=1;i<=2*n;i++)
	{
		if(colly[y[i]]==0)
		{
			colly[y[i]]=++tree;
			YY[tree]=y[i];
		}
	}
}
struct Linex
{
	int x1,x2,y;
	bool vis;//用来标记是上边还是下边
}linex[maxn];
struct Liney
{
	int x,y1,y2;
	bool vis;
}liney[maxn];
bool cmp(Linex a,Linex b)
{
	return a.y<b.y;
}
bool cmp1(Liney a,Liney b)
{
	return a.x<b.x;
}
struct ST
{
	int l,r,add;
	int len;
}st[4*maxn];
void buildtree(int id,int l,int r)
{
	st[id].l=l;
	st[id].r=r;
	if(l==r||l+1==r)
	{
		st[id].add=0;
		st[id].len=0;
		return;
	}
	int mid=(l+r)>>1;
	buildtree(2*id,l,mid);
	buildtree(2*id+1,mid,r);
	st[id].len=st[id].add=0;
}
int query(int id,int l,int r)
{
	if(st[id].add>=1)
	{
		return XX[st[id].r]-XX[st[id].l];
	}
	if(l==r||l+1==r)return 0;
	if(st[id].add)
	{
		st[2*id].add+=st[id].add;
		st[2*id+1].add+=st[id].add;
		st[id].add=0;
	}
	if(st[2*id].r>=r)
	{
		return query(2*id,l,r);
	}
	if(st[2*id+1].l<=l)
	{
		return query(2*id+1,l,r);
	}
	return query(2*id,l,st[2*id].r)+query(2*id+1,st[2*id+1].l,st[2*id+1].r);
}
void update(int id,int l,int r,int ope)
{
	if(st[id].l==l && st[id].r==r)
	{
		st[id].add+=ope;
		st[id].len=query(id,l,r);
		return;
	}
	if(st[id].add!=0)
	{
		st[2*id].add+=st[id].add;
		st[2*id+1].add+=st[id].add;
		st[id].add=0;
	}
	if(st[2*id].r>=r)
	{
		update(2*id,l,r,ope);
		st[id].len=st[2*id].len+st[2*id+1].len;
		return;
	}
	if(st[2*id+1].l<=l)
	{
		update(2*id+1,l,r,ope);
		st[id].len=st[2*id].len+st[2*id+1].len;
		return;
	}
	update(2*id,l,st[2*id].r,ope);
	update(2*id+1,st[2*id+1].l,r,ope);
	st[id].len=st[2*id].len+st[2*id+1].len;
}
int query1(int id,int l,int r)
{
	if(st[id].add>=1)
	{
		return YY[st[id].r]-YY[st[id].l];
	}
	if(l==r||l+1==r)return 0;
	if(st[id].add)
	{
		st[2*id].add+=st[id].add;
		st[2*id+1].add+=st[id].add;
		st[id].add=0;
	}
	if(st[2*id].r>=r)
	{
		return query1(2*id,l,r);
	}
	if(st[2*id+1].l<=l)
	{
		return query1(2*id+1,l,r);
	}
	return query1(2*id,l,st[2*id].r)+query1(2*id+1,st[2*id+1].l,st[2*id+1].r);
}
void update1(int id,int l,int r,int ope)
{
	if(st[id].l==l && st[id].r==r)
	{
		st[id].add+=ope;
		st[id].len=query1(id,l,r);
		return;
	}
	if(st[id].add!=0)
	{
		st[2*id].add+=st[id].add;
		st[2*id+1].add+=st[id].add;
		st[id].add=0;
	}
	if(st[2*id].r>=r)
	{
		update1(2*id,l,r,ope);
		st[id].len=st[2*id].len+st[2*id+1].len;
		return;
	}
	if(st[2*id+1].l<=l)
	{
		update1(2*id+1,l,r,ope);
		st[id].len=st[2*id].len+st[2*id+1].len;
		return;
	}
	update1(2*id,l,st[2*id].r,ope);
	update1(2*id+1,st[2*id+1].l,r,ope);
	st[id].len=st[2*id].len+st[2*id+1].len;
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		collx.clear();
		colly.clear();
		if(n==0)
		{
			printf("%d\n",0);
			continue;
		}
		int x1,y1,x2,y2;
		int kx=1,ky=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			X[kx]=x1;
			linex[kx].x1=x1;
			linex[kx].x2=x2;
			linex[kx].y=y1;
			linex[kx++].vis=1;//标记1就是下边
			X[kx]=x2;
			linex[kx].x1=x1;
			linex[kx].x2=x2;
			linex[kx].y=y2;
			linex[kx++].vis=0;//标记0就是上边
			Y[ky]=y1;
			liney[ky].y1=y1;
			liney[ky].y2=y2;
			liney[ky].x=x1;
			liney[ky++].vis=1;//标记1就是左边
			Y[ky]=y2;
			liney[ky].y1=y1;
			liney[ky].y2=y2;
			liney[ky].x=x2;
			liney[ky++].vis=0;//标记1就是左边
		}
		lisanx(X);
		buildtree(1,1,tree);
		sort(linex+1,linex+2*n+1,cmp);
		int sumlenx=0;
		int lastx=0;
		for(int i=1;i<=2*n;i++)
		{
			if(linex[i].vis)
			{
				update(1,collx[linex[i].x1],collx[linex[i].x2],1);
				sumlenx+=abs(query(1,1,tree)-lastx);
				lastx=query(1,1,tree);
			}
			else
			{
				update(1,collx[linex[i].x1],collx[linex[i].x2],-1);
				sumlenx+=abs(query(1,1,tree)-lastx);
				lastx=query(1,1,tree);
			}
		}
		lisany(Y);
		buildtree(1,1,tree);
		sort(liney+1,liney+2*n+1,cmp1);
		int sumleny=0;
		int lasty=0;
		for(int i=1;i<=2*n;i++)
		{
			if(liney[i].vis)
			{
				update1(1,colly[liney[i].y1],colly[liney[i].y2],1);
				sumleny+=abs(query1(1,1,tree)-lasty);
				lasty=query1(1,1,tree);
			}
			else
			{
				update1(1,colly[liney[i].y1],colly[liney[i].y2],-1);
				sumleny+=abs(query1(1,1,tree)-lasty);
				lasty=query1(1,1,tree);
			}
		}
		printf("%d\n",sumlenx+sumleny);
	}
	return 0;
}

/*
此题优化完毕。15MS。
只需多记录下该区间的线段数就可以只扫描一遍
也就是说此题可以分为区间合并+先扫描。
还有求线段长度也有很大改善，之前我用的是找到整段为0或者整段覆盖的方法。
其实不用，只需一个PushUp.线段经过排序，同一条线段肯定是先进后出。
进得话操作我想大家都会，出的话就看他是否为叶子，如果不是他的信息由他的
左右儿子决定。仔细揣摩PushUp.
*/
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
#define maxn 5008//矩形个数
#define lson 2*id,l,mid
#define rson 2*id+1,mid,r
int X[2*maxn],kx,k;
int binary_search(int a)
{
	int l=1,r=kx-1;
	while(l < r)
	{
		int mid=(l+r) >> 1;
		if(X[mid] >= a)
		{
			r=mid;
		}
		else l=mid + 1;
	}
	return l;
}
struct Line
{
	int x1,x2,y;
	bool vis;
}line[2*maxn];
bool cmp(Line a,Line b)
{
	return a.y<b.y;
}
struct ST
{
	int l,r,len,add;
	int rcov,lcov,line;
}st[8*maxn];
void buildtree(int id,int l,int r)
{
	st[id].l=l;
	st[id].r=r;
	if(l+1==r)
	{
		st[id].add=st[id].len=st[id].rcov=st[id].lcov=st[id].line=0;
		return;
	}
	int mid=(l+r)>>1;
	buildtree(lson);
	buildtree(rson);
	st[id].add=st[id].len=st[id].rcov=st[id].lcov=st[id].line=0;
}
void PushUp(int id)
{
	if(st[id].add >= 1)
	{
		st[id].len = X[st[id].r] - X[st[id].l];
		st[id].lcov = st[id].rcov = st[id].line = 1;
	}
	else if(st[id].l+1 == st[id].r) 
	{
		st[id].len = 0;
		st[id].lcov = st[id].rcov = st[id].line = 0;
	}
	else 
	{
		st[id].len = st[2*id].len + st[2*id+1].len;
		st[id].line = st[2*id].line + st[2*id+1].line - (st[2*id].rcov && st[2*id+1].lcov);
		st[id].rcov = st[2*id+1].rcov;
		st[id].lcov = st[2*id].lcov;
	}
}
void update(int id,int l,int r,int ope)
{
	if(st[id].l == l && st[id].r == r)
	{
		st[id].add += ope;
		if(st[id].add >= 1)
		{
			st[id].line = st[id].rcov = st[id].lcov = 1;
			st[id].len = X[r]-X[l];
		}
		else if(l+1 == r)//////// 叶子
		{
			st[id].line = st[id].rcov = st[id].lcov = 0;
			st[id].len = 0;
		}
		else 
		{
			st[id].len = st[2*id].len + st[2*id+1].len;
			st[id].rcov = st[2*id+1].rcov;
			st[id].lcov = st[2*id].lcov;
			st[id].line = st[2*id].line + st[2*id+1].line - (st[2*id].rcov && st[2*id+1].lcov);
		}
		return;
	}
	if(st[2*id].r >= r)
	{
		update(2*id,l,r,ope);
		PushUp(id);
		return;
	}
	if(st[2*id+1].l <= l)
	{
		update(2*id+1,l,r,ope);
		PushUp(id);
		return;
	}
	update(2*id,l,st[2*id].r,ope);
	update(2*id+1,st[2*id+1].l,r,ope);
	PushUp(id);
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int x1,y1,x2,y2;
		kx=1;int k=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			X[kx++]=x1;
			X[kx++]=x2;
			line[k].x1=x1;
			line[k].x2=x2;
			line[k].y=y1;
			line[k++].vis=1;
			line[k].x1=x1;
			line[k].x2=x2;
			line[k].y=y2;
			line[k++].vis=0;
		}
		sort(X+1,X+kx);
		sort(line+1,line+k,cmp);
		buildtree(1,1,kx-1);
		int sum=0,lx=0,ly=line[1].y;
		for(int i=1;i<k;i++)
		{
			if(line[i].vis)
			{
				sum+=(line[i].y-ly)*st[1].line*2;
				update(1,binary_search(line[i].x1),binary_search(line[i].x2),1);
				sum+=abs(st[1].len-lx);
				ly=line[i].y;
				lx=st[1].len;
			}
			else
			{
				sum+=(line[i].y-ly)*st[1].line*2;
				update(1,binary_search(line[i].x1),binary_search(line[i].x2),-1);
				sum+=abs(st[1].len-lx);
				ly=line[i].y;
				lx=st[1].len;
			} 
		 }
		printf("%d\n",sum);
	}
	return 0;
}