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: %I64d %I64d %I64d %I64d %I64d %I64d %I64d ", 154 ca++,ans[1],ans[2],ans[4],ans[3],ans[5],ans[6],ans[7]); 155 } 156 return 0; 157 }