POJ-1177 Picture 矩形覆盖周长并

　　题目链接：http://poj.org/problem?id=1177

　　比矩形面积并麻烦点，需要更新竖边的条数（平行于x轴扫描）。。求横边的时候，保存上一个结果，加上当前长度与上一个结果差的绝对值就行了。。。

 //STATUS:C++_AC_32MS_1416KB
 #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{
     int y,x1,x2;
     int c;
     Seg(){}
     Seg(int a,int b,int c,int d):y(a),x1(b),x2(c),c(d){}
     bool operator < (const Seg& a)const{
         return y<a.y;
     }
 }seg[N];
 bool lnod[N<<],rnod[N<<];
 int len[N<<],cnt[N<<],cnt2[N<<];
 int n,m;

 void pushup(int l,int r,int rt)
 {
     if(cnt[rt]){
         lnod[rt]=rnod[rt]=true;
         len[rt]=r-l+;
         cnt2[rt]=;
     }
     else if(l==r)cnt2[rt]=lnod[rt]=rnod[rt]=len[rt]=;
     else {
         int ls=rt<<,rs=rt<<|;
         lnod[rt]=lnod[ls],rnod[rt]=rnod[rs];
         len[rt]=len[ls]+len[rs];
         cnt2[rt]=cnt2[ls]+cnt2[rs];
         if(rnod[ls] && lnod[rs])cnt2[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 main()
 {
 //    freopen("in.txt","r",stdin);
     int i,j,k,l,r,a,b,c,d,ans,last;
     const int M=;
     while(~scanf("%d",&n))
     {
         m=;
         for(i=;i<n;i++){
             scanf("%d%d%d%d",&a,&b,&c,&d);
             a+=M,b+=M,c+=M,d+=M;
             seg[m++]=Seg(b,a,c,);
             seg[m++]=Seg(d,a,c,-);
         }
         sort(seg,seg+m);
         mem(len,);mem(cnt,),mem(cnt2,);
         mem(lnod,);mem(rnod,);
         ans=last=;
         for(i=;i<m-;i++){
             update(seg[i].x1,seg[i].x2-,seg[i].c,,,);
             ans+=cnt2[]*(seg[i+].y-seg[i].y);
             ans+=abs(len[]-last);
             last=len[];
         }
         update(seg[i].x1,seg[i].x2-,seg[i].c,,,);

         printf("%d\n",ans+abs(len[]-last));
     }
     return ;
 }