Area of Simple Polygons

poj1389:http://poj.org/problem?id=1389

题意：求矩形面积的并
题解：扫描线加线段树 同poj1389

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int maxn=;
 int num;
 struct Node{
     int  l;
     int  r;
     int tp;
     int  y;
  bool operator <(Node a) const{
             return y<a.y;
       }
 }line[*maxn];
 int  arr[*maxn];
 struct Edge{
     int left;
     int right;
     int flag;
     int  sum;
 }node1[maxn*];
 void build(int l,int r,int idx){
     node1[idx].left=l;
     node1[idx].right=r;
     if(l==r){
         node1[idx].flag=;
         node1[idx].sum=;
         return;
     }
     int mid=(l+r)/;
     build(l,mid,idx<<);
     build(mid+,r,idx<<|);
     node1[idx].sum=node1[idx<<].sum+node1[idx<<|].sum;
 }
 void update(int l,int r,int f,int idx){
     if(node1[idx].left==node1[idx].right){
           node1[idx].flag+=f;
         if(node1[idx].flag)node1[idx].sum=arr[node1[idx].right+]-arr[node1[idx].left];
         if(!node1[idx].flag)node1[idx].sum=;
         return ;
     }
     int mid=(node1[idx].left+node1[idx].right)/;
     if(mid>=r)update(l,r,f,idx<<);
     else if(mid<l)update(l,r,f,idx<<|);
     else{
         update(l,mid,f,idx<<);
         update(mid+,r,f,idx<<|);
     }
     node1[idx].sum=node1[idx<<].sum+node1[idx<<|].sum;
 }
 int binaryserach(int  x){
    int l,r,mid;
       l=,r=num+;
       while (r-l>){
             mid=(l+r)>>;
             if (arr[mid]<=x) l=mid;
                else r=mid;
       }
       return l;
 }
 int main(){
     int n;int  x1,y1,x2,y2;
     long long ans;
   while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){
       if(x1==-&&x2==-&&y1==-&&y2==-)break;
         num=;int cnt=;
         line[].l=x1;line[].r=x2;line[].y=y1;line[].tp=;
           line[].l=x1;line[].r=x2;line[].y=y2;line[].tp=-;
           arr[++num]=x1;arr[++num]=x2;
       while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){
           if(x1==-&&x2==-&&y1==-&&y2==-)break;

           line[++cnt].l=x1;line[cnt].r=x2;line[cnt].y=y1;line[cnt].tp=;
           line[++cnt].l=x1;line[cnt].r=x2;line[cnt].y=y2;line[cnt].tp=-;
           arr[++num]=x1;arr[++num]=x2;
       }

       sort(arr+,arr+num+);
         ans=;
       sort(line+,line+cnt+);
       build(,cnt,);
      for(int i=;i<=cnt;i++){
           ans+=node1[].sum*(line[i].y-line[i-].y);
           int l=binaryserach(line[i].l);
           int r=binaryserach(line[i].r)-;
            update(l,r,line[i].tp,);
       }

     printf("%I64d\n",ans);
   }
 }