HDU-1255 覆盖的面积 覆盖的矩形面积并

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1255

　　需要保存区间覆盖线>=2的线段的长度，根据情况来更新。。。

 //STATUS:C++_AC_250MS_476KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Seg{
     double y,x1,x2;
     int c;
     Seg(){}
     Seg(double a,double b,double c,int d):y(a),x1(b),x2(c),c(d){}
     bool operator < (const Seg& a)const{
         return y<a.y;
     }
 }seg[N];
 double hs[N],len[N<<][];
 int cnt[N<<];
 int T,n,m;

 void pushup(int l,int r,int rt)
 {
     if(cnt[rt])len[rt][]=hs[r+]-hs[l];
     else if(l==r)len[rt][]=;
     else len[rt][]=len[rt<<][]+len[rt<<|][];
     if(cnt[rt]>=)len[rt][]=len[rt][];
     else if(cnt[rt]==)len[rt][]=len[rt<<][]+len[rt<<|][];
     else if(l==r)len[rt][]=;
     else len[rt][]=len[rt<<][]+len[rt<<|][];
 }

 void update(int a,int b,int c,int l,int r,int rt)
 {
     if(a<=l && r<=b){
         cnt[rt]+=c;
         pushup(l,r,rt);
         return;
     }
     int mid=(l+r)>>;
     if(a<=mid)update(a,b,c,lson);
     if(b>mid)update(a,b,c,rson);
     pushup(l,r,rt);
 }

 int binary(int l,int r,double tar)
 {
     int mid;
     while(l<r){
         mid=(l+r)>>;
         if(hs[mid]==tar)return mid;
         else if(hs[mid]>tar)r=mid;
         else l=mid+;
     }
     return -;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,k,l,r;
     double ans,a,b,c,d;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         m=;
         for(i=;i<n;i++){
             scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
             hs[m]=a;
             seg[m++]=Seg(b,a,c,);
             hs[m]=c;
             seg[m++]=Seg(d,a,c,-);
         }
         sort(hs,hs+m);
         sort(seg,seg+m);
         for(i=,k=;i<m;i++)
             if(hs[i]!=hs[k])hs[++k]=hs[i];
         mem(len,);mem(cnt,);
         ans=;
         for(i=;i<m-;i++){
             l=binary(,k+,seg[i].x1);
             r=binary(,k+,seg[i].x2)-;
             if(l<=r)update(l,r,seg[i].c,,k,);
             ans+=len[][]*(seg[i+].y-seg[i].y);
         }

         printf("%.2lf\n",ans);
     }
     return ;
 }