POJ-1151 Atlantis 矩形面积并

  题目链接:http://poj.org/problem?id=1151

  扫描线+离散+线段树,线段树每个节点保存的是离散后节点右边的线段。

  1 //STATUS:C++_AC_16MS_208KB
  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=210;
 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     double y,x1,x2;
 59     int c;
 60     Seg(){}
 61     Seg(double a,double b,double 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 double hs[N],len[N<<2];
 67 int cnt[N<<2];
 68 int n,m;
 69 
 70 void pushup(int l,int r,int rt)
 71 {
 72     if(cnt[rt])len[rt]=hs[r+1]-hs[l];
 73     else if(l==r)len[rt]=0;
 74     else len[rt]=len[rt<<1]+len[rt<<1|1];
 75 }
 76 
 77 void update(int a,int b,int c,int l,int r,int rt)
 78 {
 79     if(a<=l && r<=b){
 80         cnt[rt]+=c;
 81         pushup(l,r,rt);
 82         return;
 83     }
 84     int mid=(l+r)>>1;
 85     if(a<=mid)update(a,b,c,lson);
 86     if(b>mid)update(a,b,c,rson);
 87     pushup(l,r,rt);
 88 }
 89 
 90 int binary(int l,int r,double tar)
 91 {
 92     int mid;
 93     while(l<r){
 94         mid=(l+r)>>1;
 95         if(hs[mid]==tar)return mid;
 96         else if(hs[mid]>tar)r=mid;
 97         else l=mid+1;
 98     }
 99     return -1;
100 }
101 
102 int main()
103 {
104  //   freopen("in.txt","r",stdin);
105     int i,j,k,ca=1,l,r;
106     double ans,a,b,c,d;
107     while(~scanf("%d",&n) && n)
108     {
109         m=0;
110         for(i=0;i<n;i++){
111             scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
112             hs[m]=a;
113             seg[m++]=Seg(b,a,c,1);
114             hs[m]=c;
115             seg[m++]=Seg(d,a,c,-1);
116         }
117         sort(hs,hs+m);
118         sort(seg,seg+m);
119         for(i=1,k=0;i<m;i++)
120             if(hs[i]!=hs[k])hs[++k]=hs[i];
121         mem(len,0);mem(cnt,0);
122         ans=0;
123         for(i=0;i<m-1;i++){
124             l=binary(0,k+1,seg[i].x1);
125             r=binary(0,k+1,seg[i].x2)-1;
126             if(l<=r)update(l,r,seg[i].c,0,k,1);
127             ans+=len[1]*(seg[i+1].y-seg[i].y);
128         }
129 
130         printf("Test case #%d
Total explored area: %.2lf

",ca++,ans);
131     }
132     return 0;
133 }
原文地址:https://www.cnblogs.com/zhsl/p/3196636.html