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。。。

  1 //STATUS:C++_AC_32MS_1416KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //using namespace __gnu_cxx;

 25 //define

 26 #define pii pair<int,int>

 27 #define mem(a,b) memset(a,b,sizeof(a))

 28 #define lson l,mid,rt<<1

 29 #define rson mid+1,r,rt<<1|1

 30 #define PI acos(-1.0)

 31 //typedef

 32 typedef __int64 LL;

 33 typedef unsigned __int64 ULL;

 34 //const

 35 const int N=20010;

 36 const int INF=0x3f3f3f3f;

 37 const int MOD=100000,STA=8000010;

 38 const LL LNF=1LL<<60;

 39 const double EPS=1e-8;

 40 const double OO=1e15;

 41 const int dx[4]={-1,0,1,0};

 42 const int dy[4]={0,1,0,-1};

 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 44 //Daily Use ...

 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 55 //End

 56 

 57 struct Seg{

 58     int y,x1,x2;

 59     int c;

 60     Seg(){}

 61     Seg(int a,int b,int c,int d):y(a),x1(b),x2(c),c(d){}

 62     bool operator < (const Seg& a)const{

 63         return y<a.y;

 64     }

 65 }seg[N];

 66 bool lnod[N<<2],rnod[N<<2];

 67 int len[N<<2],cnt[N<<2],cnt2[N<<2];

 68 int n,m;

 69 

 70 void pushup(int l,int r,int rt)

 71 {

 72     if(cnt[rt]){

 73         lnod[rt]=rnod[rt]=true;

 74         len[rt]=r-l+1;

 75         cnt2[rt]=2;

 76     }

 77     else if(l==r)cnt2[rt]=lnod[rt]=rnod[rt]=len[rt]=0;

 78     else {

 79         int ls=rt<<1,rs=rt<<1|1;

 80         lnod[rt]=lnod[ls],rnod[rt]=rnod[rs];

 81         len[rt]=len[ls]+len[rs];

 82         cnt2[rt]=cnt2[ls]+cnt2[rs];

 83         if(rnod[ls] && lnod[rs])cnt2[rt]-=2;

 84     }

 85 }

 86 

 87 void update(int a,int b,int c,int l,int r,int rt)

 88 {

 89     if(a<=l && r<=b){

 90         cnt[rt]+=c;

 91         pushup(l,r,rt);

 92         return;

 93     }

 94     int mid=(l+r)>>1;

 95     if(a<=mid)update(a,b,c,lson);

 96     if(b>mid)update(a,b,c,rson);

 97     pushup(l,r,rt);

 98 }

 99 

100 int main()

101 {

102 //    freopen("in.txt","r",stdin);

103     int i,j,k,l,r,a,b,c,d,ans,last;

104     const int M=10000;

105     while(~scanf("%d",&n))

106     {

107         m=0;

108         for(i=0;i<n;i++){

109             scanf("%d%d%d%d",&a,&b,&c,&d);

110             a+=M,b+=M,c+=M,d+=M;

111             seg[m++]=Seg(b,a,c,1);

112             seg[m++]=Seg(d,a,c,-1);

113         }

114         sort(seg,seg+m);

115         mem(len,0);mem(cnt,0),mem(cnt2,0);

116         mem(lnod,0);mem(rnod,0);

117         ans=last=0;

118         for(i=0;i<m-1;i++){

119             update(seg[i].x1,seg[i].x2-1,seg[i].c,0,20000,1);

120             ans+=cnt2[1]*(seg[i+1].y-seg[i].y);

121             ans+=abs(len[1]-last);

122             last=len[1];

123         }

124         update(seg[i].x1,seg[i].x2-1,seg[i].c,0,20000,1);

125 

126         printf("%d\n",ans+abs(len[1]-last));

127     }

128     return 0;

129 }