| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
Re:线段树+离散化+扫描线In Reply To:线段树+离散化+扫描线 Posted by:ecjtu_yuweiwei at 2014-07-29 14:58:50 > #include<iostream>
> #include<queue>
> #include<cstdio>]o9ikas
> #include<algorithm>
> #include<cstring>
> #include<iomanip>
> #include<map>
> #include<cstdlib>
> #include<cmath>
> #include<vector>
> #define LL long long
> #define IT __int64
> #define zero(x) fabs(x)<eps
> #define mm(a,b) memset(a,b,sizeof(a))
> const int INF=0x7fffffff;
> const double inf=1e8;
> const double eps=1e-10;
> const double PI=acos(-1.0);
> const int Max=201;
> using namespace std;
> int sign(double x)
> {
> return (x>eps)-(x<-eps);
> }
> struct Node
> {
> int left;
> int right;//线段树的左右整点
> int flag;//记录重叠情况,大于零说明没有重叠
> double cnt;//记录实际的长度
> double lf;//左边端点真实的浮点数
> double rf;//右边端点真是的浮点数
> }segTree[Max<<2];
> struct Line
> {
> double x;
> double y1;
> double y2;
> int ok;
> }line[Max];
> double y[Max];//记录y坐标的数组
> //把一段段平行于y轴的线段表示成数组
> //x是线段的x坐标,y1,y2线段对应的下端点和上端点的坐标
> //一个矩形 ,左边的那条边ok为1,右边的为-1
> //用来记录重叠情况,可以根据node节点中的flag这个来计算
> bool cmp(Line u,Line v)//sort排序
> {
> return u.x<v.x;
> }
> void Build_Tree(int t,int left,int right)
> {
> segTree[t].left=left;
> segTree[t].right=right;
> segTree[t].lf=y[left];
> segTree[t].rf=y[right];
> if((left+1)==right) return;
> int mid=(left+right)>>1;
> Build_Tree(t<<1,left,mid);
> Build_Tree(t<<1|1,mid,right);//递归构造线段树,这里mid不能+1因为如果+1那么t<<1的右孩子和t<<1|1的左孩子不能产生联系最终更新父节点就是错误的数据
> }
> void Calen(int t)//计算长度
> {
> if(segTree[t].flag>0)//如果没有重叠
> {
> segTree[t].cnt=segTree[t].rf-segTree[t].lf;
> }
> else if(segTree[t].left+1==segTree[t].right)//如果重叠了
> {
> segTree[t].cnt=0;
> }
> else
> {
> segTree[t].cnt=segTree[t<<1].cnt+segTree[t<<1|1].cnt;
> }
> }
> void Update(int t,Line L)//加入线段L,后更新线段树
> {
> if(L.y1==segTree[t].lf&&L.y2==segTree[t].rf)
> {
> segTree[t].flag+=L.ok;
> Calen(t);
> }
> else if(L.y2<=segTree[t<<1].rf)
> {
> Update(t<<1,L);
> }
> else if(L.y1>=segTree[t<<1|1].lf)
> {
> Update(t<<1|1,L);
> }
> else
> {
> Line temp;
> temp=L;
> temp.y2=segTree[t<<1].rf;
> Update(t<<1,temp);
> temp=L;
> temp.y1=segTree[t<<1|1].lf;
> Update(t<<1|1,temp);
> }
> Calen(t);
> }
> void Init(int &m,double x1,double y1,double x2,double y2)
> {
> line[m].x=x1;
> line[m].y1=y1;
> line[m].y2=y2;
> line[m].ok=1;
> y[m]=y1;
> m++;
> line[m].x=x2;
> line[m].y1=y1;
> line[m].y2=y2;
> line[m].ok=-1;
> y[m]=y2;
> m++;
> }
> int main()
> {
> int n,m,i,j,Case;
> double x1,y1,x2,y2,area;
> Case=1;
> while(cin>>n&&n)
> {
> m=1;
> for(i=1;i<=n;i++)
> {
> cin>>x1>>y1>>x2>>y2;
> Init(m,x1,y1,x2,y2);
> }
> sort(line+1,line+m,cmp);
> sort(y+1,y+m);
> Build_Tree(1,1,m-1);
> Update(1,line[1]);
> area=0;
> for(i=2;i<m;i++)
> {
> //cout<<segTree[1].cnt<<" "<<line[i-1].x<<" "<<line[i].x<<endl;
> area+=segTree[1].cnt*(line[i].x-line[i-1].x);
> Update(1,line[i]);
> }
> cout<<"Test case #"<<Case++<<endl;
> cout<<"Total explored area: "<<setprecision(2)<<setiosflags(ios::fixed)<<area<<endl<<endl;
> }
> return 0;
> }
Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator