HDU-1255 覆盖的面积 覆盖的矩形面积并

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1255

　　需要保存区间覆盖线>=2的线段的长度，根据情况来更新。。。

  1 //STATUS:C++_AC_250MS_476KB

  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=2010;

 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][2];

 67 int cnt[N<<2];

 68 int T,n,m;

 69 

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

 71 {

 72     if(cnt[rt])len[rt][0]=hs[r+1]-hs[l];

 73     else if(l==r)len[rt][0]=0;

 74     else len[rt][0]=len[rt<<1][0]+len[rt<<1|1][0];

 75     if(cnt[rt]>=2)len[rt][1]=len[rt][0];

 76     else if(cnt[rt]==1)len[rt][1]=len[rt<<1][0]+len[rt<<1|1][0];

 77     else if(l==r)len[rt][1]=0;

 78     else len[rt][1]=len[rt<<1][1]+len[rt<<1|1][1];

 79 }

 80 

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

 82 {

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

 84         cnt[rt]+=c;

 85         pushup(l,r,rt);

 86         return;

 87     }

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

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

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

 91     pushup(l,r,rt);

 92 }

 93 

 94 int binary(int l,int r,double tar)

 95 {

 96     int mid;

 97     while(l<r){

 98         mid=(l+r)>>1;

 99         if(hs[mid]==tar)return mid;

100         else if(hs[mid]>tar)r=mid;

101         else l=mid+1;

102     }

103     return -1;

104 }

105 

106 int main()

107 {

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

109     int i,j,k,l,r;

110     double ans,a,b,c,d;

111     scanf("%d",&T);

112     while(T--)

113     {

114         scanf("%d",&n);

115         m=0;

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

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

118             hs[m]=a;

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

120             hs[m]=c;

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

122         }

123         sort(hs,hs+m);

124         sort(seg,seg+m);

125         for(i=1,k=0;i<m;i++)

126             if(hs[i]!=hs[k])hs[++k]=hs[i];

127         mem(len,0);mem(cnt,0);

128         ans=0;

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

130             l=binary(0,k+1,seg[i].x1);

131             r=binary(0,k+1,seg[i].x2)-1;

132             if(l<=r)update(l,r,seg[i].c,0,k,1);

133             ans+=len[1][1]*(seg[i+1].y-seg[i].y);

134         }

135 

136         printf("%.2lf\n",ans);

137     }

138     return 0;

139 }