uva 11983 Weird Advertisement(扫描线)

题意:给出N个矩形,求被这些矩形覆盖K次以上的区域面积。

做过hdu 3642 Get The Treasure和hdu 1255 覆盖的面积这道题应该不难才是。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(a,b) (a+((b-a)>>1))
typedef long long LL;
const int N=60005;

struct Line
{
	int x,y1,y2,flag;
	Line(){}
	Line(int a,int b,int c,int d)
	{ x=a;y1=b;y2=c;flag=d; }
	bool operator<(const Line&b)const
	{ return x<b.x; }
};
struct node
{
	int lft,rht;
	int len[12],flag;
	int mid(){return MID(lft,rht);}
	void init(){memset(len,0,sizeof(len));}
};

int n,k;
vector<int> y;
vector<Line> line;
map<int,int> H;

struct Segtree
{
	node tree[N*4];
	void calu(int ind)
	{
		if(tree[ind].flag>=k)
		{
			int tmp=tree[ind].len[0];
			tree[ind].init();
			tree[ind].len[k]=tree[ind].len[0]=tmp;
		}
		else if(tree[ind].flag>0)
		{
			int sum=0,cov=tree[ind].flag;
			for(int i=1;i<=k;i++) tree[ind].len[i]=0;
			tree[ind].len[cov]=tree[ind].len[0];
			if(tree[ind].lft+1==tree[ind].rht) return;

			for(int i=1;i<=k;i++)
			{
				if(i+cov>=k) tree[ind].len[k]+=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];
				else tree[ind].len[i+cov]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];
			}
			for(int i=cov+1;i<=k;i++) sum+=tree[ind].len[i];
			tree[ind].len[cov]-=sum;
		}
		else 
		{
			for(int i=1;i<=k;i++) tree[ind].len[i]=0;
			if(tree[ind].lft+1==tree[ind].rht) return; 
			for(int i=1;i<=k;i++)
				tree[ind].len[i]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];
		}
	}
	void build(int lft,int rht,int ind)
	{
		tree[ind].lft=lft;	tree[ind].rht=rht;
		tree[ind].init(); 	tree[ind].flag=0;
		tree[ind].len[0]=y[rht]-y[lft];
		if(lft+1!=rht)
		{
			int mid=tree[ind].mid();
			build(lft,mid,LL(ind));
			build(mid,rht,RR(ind));
		}
	}
	void updata(int st,int ed,int ind,int valu)
	{
		int lft=tree[ind].lft,rht=tree[ind].rht;
		if(st<=lft&&rht<=ed) tree[ind].flag+=valu;
		else 
		{
			int mid=tree[ind].mid();
			if(st<mid) updata(st,ed,LL(ind),valu);
			if(ed>mid) updata(st,ed,RR(ind),valu);
		}
		calu(ind);
	}
}seg;
int main()
{
	int t,t_cnt=0;
	scanf("%d",&t);
	while(t--)
	{
		y.clear(); H.clear(); line.clear();

		scanf("%d%d",&n,&k);
		for(int i=0;i<n;i++)
		{
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			x2++;y2++;
			line.push_back(Line(x1,y1,y2,1));
			line.push_back(Line(x2,y1,y2,-1));
			y.push_back(y1); y.push_back(y2);
		}

		sort(line.begin(),line.end());
		sort(y.begin(),y.end());
		y.erase(unique(y.begin(),y.end()),y.end());
		for(int i=0;i<(int)y.size();i++) H[y[i]]=i;

		seg.build(0,(int)y.size()-1,1);

		LL res=0;
		for(int i=0;i<(int)line.size();i++)
		{
			if(i!=0) res+=(LL)(line[i].x-line[i-1].x)*seg.tree[1].len[k];
			seg.updata(H[line[i].y1],H[line[i].y2],1,line[i].flag);
		}
		printf("Case %d: %lld\n",++t_cnt,res);
	}
	return 0;
}


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