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\nTotal explored area: %.2lf\n\n",ca++,ans); 131 } 132 return 0; 133 }