HDU-4419 Colourful Rectangle 矩形多面积并

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4419

　　利用二进制，R为1、G为2、B为4，然后通过异或运算可以得到其它组合颜色。建立7颗线段树，每颗线段树保存每种颜色的长度。。。

  1 //STATUS:C++_AC_203MS_4780KB

  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,col;

 60     Seg(){}

 61     Seg(int a,int b,int c,int d,int color):y(a),x1(b),x2(c),c(d),col(color){}

 62     bool operator < (const Seg& a)const{

 63         return y<a.y;

 64     }

 65 }seg[N];

 66 int hs[N],len[N<<2][8];

 67 int cnt[N<<2][5];

 68 LL ans[8];

 69 int T,n,m;

 70 

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

 72 {

 73     int i,col=((cnt[rt][1]?1:0) | (cnt[rt][2]?2:0) | (cnt[rt][4]?4:0));

 74     if(col){

 75         int ls=rt<<1,rs=rt<<1|1;

 76         mem(len[rt],0);

 77         len[rt][col]=hs[r+1]-hs[l];

 78         i=1;

 79         for(i=1;i<=7;i++){

 80             if(col==(col|i))continue;

 81             int t=len[ls][i]+len[rs][i];

 82             len[rt][col|i]+=t;

 83             len[rt][col]-=t;

 84         }

 85     }

 86     else if(l==r)mem(len[rt],0);

 87     else {

 88         int ls=rt<<1,rs=rt<<1|1;

 89         for(i=1;i<=7;i++){

 90             len[rt][i]=len[ls][i]+len[rs][i];

 91         }

 92     }

 93 }

 94 

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

 96 {

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

 98         cnt[rt][col]+=c;

 99         pushup(l,r,rt);

100         return;

101     }

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

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

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

105     pushup(l,r,rt);

106 }

107 

108 int binary(int l,int r,int tar)

109 {

110     int mid;

111     while(l<r){

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

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

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

115         else l=mid+1;

116     }

117     return -1;

118 }

119 

120 int main()

121 {

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

123     int key[130];

124     key['R']=1,key['G']=2,key['B']=4;

125     int i,j,k,l,r,ca=1,a,b,c,d;

126     char s[2];

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

128     while(T--)

129     {

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

131         m=0;

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

133             scanf("%s%d%d%d%d",s,&a,&b,&c,&d);

134             hs[m]=a;

135             seg[m++]=Seg(b,a,c,1,key[s[0]]);

136             hs[m]=c;

137             seg[m++]=Seg(d,a,c,-1,key[s[0]]);

138         }

139         sort(hs,hs+m);

140         sort(seg,seg+m);

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

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

143         mem(len,0);mem(cnt,0);mem(ans,0);

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

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

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

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

148             for(j=1;j<=7;j++){

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

150             }

151         }

152 

153         printf("Case %d:\n%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n",

154                ca++,ans[1],ans[2],ans[4],ans[3],ans[5],ans[6],ans[7]);

155     }

156     return 0;

157 }