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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <limits> 8 #include <vector> 9 #include <stack> 10 #include <queue> 11 #include <set> 12 #include <map> 13 #define lowbit(x) ( x&(-x) ) 14 #define pi 3.141592653589793 15 #define e 2.718281828459045 16 #define INF 0x3f3f3f3f 17 #define HalF (l + r)>>1 18 #define lsn rt<<1 19 #define rsn rt<<1|1 20 #define Lson lsn, l, mid 21 #define Rson rsn, mid+1, r 22 #define QL Lson, ql, qr 23 #define QR Rson, ql, qr 24 #define myself rt, l, r 25 using namespace std; 26 typedef unsigned long long ull; 27 typedef long long ll; 28 const int maxN = 1e4 + 7; 29 int N, tot, _UP; 30 ll X[maxN<<1], ans[8]; 31 struct node 32 { 33 ll lx, rx, y; 34 int typ, val; 35 node(ll a=0, ll b=0, ll c=0, int f=0, int d=0):lx(a), rx(b), y(c), typ(f), val(d) {} 36 }line[maxN<<1]; 37 bool cmp(node e1, node e2) { return e1.y < e2.y; } 38 struct Tree 39 { 40 int siz[4]; 41 ll col[8]; 42 void clear() { memset(siz, 0, sizeof(siz)); memset(col, 0, sizeof(col)); } 43 }t[maxN<<3]; 44 inline void pushup(int rt, int l, int r) 45 { 46 memset(t[rt].col, 0, sizeof(t[rt].col)); 47 int state = 0; 48 for(int i=1; i<=3; i++) if(t[rt].siz[i]) state |= (1<<(i - 1)); 49 if(l == r) t[rt].col[state] = X[r + 1] - X[l]; 50 else for(int i=0; i<=7; i++) { t[rt].col[i|state] += t[lsn].col[i] + t[rsn].col[i]; } 51 } 52 inline void buildTree(int rt, int l, int r) 53 { 54 t[rt].clear(); 55 if(l == r) { t[rt].col[0] = X[r + 1] - X[l]; return; } 56 int mid = HalF; 57 buildTree(Lson); 58 buildTree(Rson); 59 t[rt].col[0] = t[lsn].col[0] + t[rsn].col[0]; 60 } 61 inline void update(int rt, int l, int r, int ql, int qr, int typ, int val) 62 { 63 if(ql <= l && qr >= r) 64 { 65 t[rt].siz[typ] += val; 66 pushup(myself); 67 return; 68 } 69 int mid = HalF; 70 if(qr <= mid) update(QL, typ, val); 71 else if(ql > mid) update(QR, typ, val); 72 else { update(QL, typ, val); update(QR, typ, val); } 73 pushup(myself); 74 } 75 inline void init() 76 { 77 tot = 0; 78 memset(ans, 0, sizeof(ans)); 79 } 80 char s[5]; 81 int main() 82 { 83 int T; scanf("%d", &T); 84 for(int Cas=1; Cas<=T; Cas++) 85 { 86 scanf("%d", &N); 87 init(); 88 ll lx, ly, rx, ry; 89 for(int i=1, tmp; i<=N; i++) 90 { 91 scanf("%s%lld%lld%lld%lld", s, &lx, &ly, &rx, &ry); 92 if(s[0] == 'R') tmp = 1; 93 else if(s[0] == 'G') tmp = 2; 94 else tmp = 3; 95 line[++tot] = node(lx, rx, ly, tmp, 1); 96 X[tot] = lx; 97 line[++tot] = node(lx, rx, ry, tmp, -1); 98 X[tot] = rx; 99 } 100 sort(X + 1, X + tot + 1); 101 sort(line + 1, line + tot + 1, cmp); 102 _UP = (int)(unique(X + 1, X + tot + 1) - X - 1); 103 buildTree(1, 1, _UP); 104 for(int i=1, l, r; i<tot; i++) 105 { 106 l =(int)(lower_bound(X + 1, X + _UP + 1, line[i].lx) - X); 107 r = (int)(lower_bound(X + 1, X + _UP + 1, line[i].rx) - X - 1); 108 update(1, 1, _UP, l, r, line[i].typ, line[i].val); 109 for(int col=1; col<=7; col++) ans[col] += (line[i + 1].y - line[i].y) * t[1].col[col]; 110 } 111 swap(ans[3], ans[4]); 112 printf("Case %d:\n", Cas); 113 for(int i=1; i<=7; i++) printf("%lld\n", ans[i]); 114 } 115 return 0; 116 }
