Colourful Rectangle【扫描线】

题目链接

很明显的可以发现是一个扫描线的问题，但是怎么处理区域呢，发现只有三种颜色，也就是最多也就是7种状态，那么我们可以进行一个状态压缩即可。

但是，在向上pushup的时候，存在我们要以子树的状态来向上推，也就意味着一开始的目前节点是要去清空的，然后再去更新其覆盖的线长。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <string>
 #include <cstring>
 #include <algorithm>
 #include <limits>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #define lowbit(x) ( x&(-x) )
 #define pi 3.141592653589793
 #define e 2.718281828459045
 #define INF 0x3f3f3f3f
 #define HalF (l + r)>>1
 #define lsn rt<<1
 #define rsn rt<<1|1
 #define Lson lsn, l, mid
 #define Rson rsn, mid+1, r
 #define QL Lson, ql, qr
 #define QR Rson, ql, qr
 #define myself rt, l, r
 using namespace std;
 typedef unsigned long long ull;
 typedef long long ll;
 const int maxN = 1e4 + ;
 int N, tot, _UP;
 ll X[maxN<<], ans[];
 struct node
 {
     ll lx, rx, y;
     int typ, val;
     node(ll a=, ll b=, ll c=, int f=, int d=):lx(a), rx(b), y(c), typ(f), val(d) {}
 }line[maxN<<];
 bool cmp(node e1, node e2) { return e1.y < e2.y; }
 struct Tree
 {
     int siz[];
     ll col[];
     void clear() { memset(siz, , sizeof(siz)); memset(col, , sizeof(col)); }
 }t[maxN<<];
 inline void pushup(int rt, int l, int r)
 {
     memset(t[rt].col, , sizeof(t[rt].col));
     int state = ;
     for(int i=; i<=; i++) if(t[rt].siz[i]) state |= (<<(i - ));
     if(l == r) t[rt].col[state] = X[r + ] - X[l];
     else for(int i=; i<=; i++) { t[rt].col[i|state] += t[lsn].col[i] + t[rsn].col[i]; }
 }
 inline void buildTree(int rt, int l, int r)
 {
     t[rt].clear();
     if(l == r) { t[rt].col[] = X[r + ] - X[l]; return; }
     int mid = HalF;
     buildTree(Lson);
     buildTree(Rson);
     t[rt].col[] = t[lsn].col[] + t[rsn].col[];
 }
 inline void update(int rt, int l, int r, int ql, int qr, int typ, int val)
 {
     if(ql <= l && qr >= r)
     {
         t[rt].siz[typ] += val;
         pushup(myself);
         return;
     }
     int mid = HalF;
     if(qr <= mid) update(QL, typ, val);
     else if(ql > mid) update(QR, typ, val);
     else { update(QL, typ, val); update(QR, typ, val); }
     pushup(myself);
 }
 inline void init()
 {
     tot = ;
     memset(ans, , sizeof(ans));
 }
 char s[];
 int main()
 {
     int T;  scanf("%d", &T);
     for(int Cas=; Cas<=T; Cas++)
     {
         scanf("%d", &N);
         init();
         ll lx, ly, rx, ry;
         for(int i=, tmp; i<=N; i++)
         {
             scanf("%s%lld%lld%lld%lld", s, &lx, &ly, &rx, &ry);
             if(s[] == 'R') tmp = ;
             else if(s[] == 'G') tmp = ;
             else tmp = ;
             line[++tot] = node(lx, rx, ly, tmp, );
             X[tot] = lx;
             line[++tot] = node(lx, rx, ry, tmp, -);
             X[tot] = rx;
         }
         sort(X + , X + tot + );
         sort(line + , line + tot + , cmp);
         _UP = (int)(unique(X + , X + tot + ) - X - );
         buildTree(, , _UP);
         for(int i=, l, r; i<tot; i++)
         {
             l  =(int)(lower_bound(X + , X + _UP + , line[i].lx) - X);
             r = (int)(lower_bound(X + , X + _UP + , line[i].rx) - X - );
             update(, , _UP, l, r, line[i].typ, line[i].val);
             for(int col=; col<=; col++) ans[col] += (line[i + ].y - line[i].y) * t[].col[col];
         }
         swap(ans[], ans[]);
         printf("Case %d:\n", Cas);
         for(int i=; i<=; i++) printf("%lld\n", ans[i]);
     }
     return ;
 }