[HDU 4419] Colourful Rectangle (扫描线 矩形面积并)

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

题目大意：比矩形面积并多了颜色，问染成的每种颜色的面积。

矩形面积并的扫描线维护的是长度，这道题就是维护每个颜色的长度，写起来很蛋疼。

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;
 typedef pair<int,int> PII;
 typedef long long LL;

 struct Node{
     int l,r,h,d,c;
     bool operator<(const Node& a) const{
         return h<a.h;
     }
     Node(int L=,int R=,int H=,int D=,int C=):l(L),r(R),h(H),d(D),c(C){}
 };
 const int MAX_N = ;
 int T,n;
 Node t[MAX_N<<];
 int dsum[MAX_N<<][],b[MAX_N<<];
 LL sum[];
 int scol[MAX_N<<],len[MAX_N<<][];

 void push_up(int idx,int l,int r){
     int state = (dsum[idx][]>?:)|(dsum[idx][]>?:)|(dsum[idx][]>?:);
     if( state ) {
         memset(len[idx],,sizeof(len[idx]));
         len[idx][state] = b[r+] - b[l];
         for(int i=;i<;i++){
             if( state!=(state|i) ){
                 int tmp = len[idx<<][i]+len[idx<<|][i];
                 len[idx][state|i] += tmp;
                 len[idx][state] -= tmp;
             }
         }
     } else if(l!=r){
         for(int i=;i<;i++){
             len[idx][i] = len[idx<<][i] + len[idx<<|][i];
         }
     } else {
         memset(len[idx],,sizeof(len[idx]));
     }
 }

 void update(int L,int R,int x,int c,int idx,int l,int r){
     if( L<=l&&R>=r ){
         dsum[idx][c] += x;
     } else {
         int m = l+r>>;
         if( L<=m ) update(L,R,x,c,idx<<,l,m);
         if( R>m ) update(L,R,x,c,idx<<|,m+,r);
     }
     push_up(idx,l,r);
 }

 int main(){
     scanf("%d",&T);
     int kase = ;
     while(T--){
         memset(dsum,,sizeof(dsum));
         memset(len,,sizeof(len));
         memset(scol,,sizeof(scol));
         scanf("%d",&n);
         char col[];
         int x1,y1,x2,y2;
         int ptr =  , ptrb = ;
         for(int i=;i<n;i++){
             scanf("%s%d%d%d%d",col,&x1,&y1,&x2,&y2);
             int cc;
             if( col[]=='R' ) cc=;
             else if( col[]=='G' ) cc = ;
             else if( col[]=='B' ) cc = ;
             t[ptr++] = Node(x1,x2,y1,,cc);
             t[ptr++] = Node(x1,x2,y2,-,cc);
             b[ptrb++] = x1;
             b[ptrb++] = x2;
         }
         sort(t,t+ptr);
         sort(b,b+ptrb);
         int ub = unique(b,b+ptrb) - b;
         memset(sum,,sizeof(sum));

         for(int i=;i<ptr-;i++){
             int l = lower_bound(b,b+ub,t[i].l) - b;
             int r = lower_bound(b,b+ub,t[i].r) - b - ;
             update(l,r,t[i].d,t[i].c,,,ub-);
             if( t[i].h!=t[i+].h ){
                 for(int j=;j<;j++){
                     sum[j] += (LL)(t[i+].h-t[i].h)*(LL)(len[][j]);
                 }
             }
         }
         printf("Case %d:\n",kase++);
         printf("%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n%I64d\n",sum[],sum[],sum[],sum[],sum[],sum[],sum[]);
     }
     return ;
 }