• 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 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/3196636.html
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