POJ-1177 Picture 矩形覆盖周长并

　　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 ;

 }
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POJ-1177 Picture 矩形覆盖周长并

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