首页 > 技术文章 > POJ-1177 Picture 矩形覆盖周长并

zhsl 2013-07-17 20:35 原文

  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 }

 

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