妈蛋真简单。。。
离线版的可以乱搞。。。貌似是BZOJ的2683
在线的话。。一开始想到了BIT套平衡树。。。空间复杂度$O(n * logm)$非常开心的MLE了。。。QAQ
于是写KD树，空间复杂度$O(n)$，时间复杂度$O(msqrt(m))$，于是就过了QAQQQ
看错题了非常开心。。。一开始读入的终止条件写成了：
for (i = 1; i <= n; ++i)
于是再见了QAQQQ
1 /************************************************************** 2 Problem: 4066 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:42812 ms 7 Memory:9408 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 const int M = 2e5 + 5; 16 17 int n; 18 ll ans; 19 20 inline int read(int f = 1) { 21 static int x; 22 static char ch; 23 x = 0, ch = getchar(); 24 while (ch < '0' || '9' < ch) 25 ch = getchar(); 26 while ('0' <= ch && ch <= '9') 27 x = x * 10 + ch - '0', ch = getchar(); 28 if (f) x ^= ans; 29 return x; 30 } 31 32 struct point { 33 int x[2], v; 34 35 int& operator [] (int i) { 36 return x[i]; 37 } 38 inline bool operator == (const point &p) const { 39 return x[0] == p.x[0] && x[1] == p.x[1]; 40 } 41 inline void get() { 42 x[0] = read(), x[1] = read(), v = read(); 43 } 44 }; 45 46 struct KD_tree { 47 KD_tree *son[2]; 48 point p; 49 int mn[2], mx[2]; 50 ll sum; 51 52 KD_tree(point _p) { 53 son[0] = son[1] = NULL, p.v = sum = _p.v; 54 p[0] = mn[0] = mx[0] = _p[0], p[1] = mn[1] = mx[1] = _p[1]; 55 } 56 KD_tree() {} 57 58 inline void* operator new(size_t, point _p) { 59 static KD_tree mempool[M], *c = mempool; 60 *c = KD_tree(_p); 61 return c++; 62 } 63 64 inline void update() { 65 static int i; 66 for (i = 0; i < 2; ++i) { 67 mn[i] = mx[i] = p[i]; 68 if (son[0]) { 69 mn[i] = min(mn[i], son[0] -> mn[i]); 70 mx[i] = max(mx[i], son[0] -> mx[i]); 71 } 72 if (son[1]) { 73 mn[i] = min(mn[i], son[1] -> mn[i]); 74 mx[i] = max(mx[i], son[1] -> mx[i]); 75 } 76 } 77 sum = p.v; 78 if (son[0]) sum += son[0] -> sum; 79 if (son[1]) sum += son[1] -> sum; 80 } 81 82 void insert(point _p, int dep) { 83 if (p == _p) { 84 p.v += _p.v, sum += _p.v; 85 return; 86 } 87 bool d = _p[dep] < p[dep]; 88 if (!son[d]) son[d] = new(_p)KD_tree; 89 else son[d] -> insert(_p, !dep); 90 update(); 91 } 92 93 inline bool in(int x1, int y1, int x2, int y2, int X1, int Y1, int X2, int Y2) { 94 return x1 <= X1 && X2 <= x2 && y1 <= Y1 && Y2 <= y2; 95 } 96 97 inline bool out(int x1, int y1, int x2, int y2, int X1, int Y1, int X2, int Y2) { 98 return X2 < x1 || x2 < X1 || Y2 < y1 || y2 < Y1; 99 } 100 101 ll query(int x1, int y1, int x2, int y2) { 102 ll res = 0; 103 if (in(x1, y1, x2, y2, mn[0], mn[1], mx[0], mx[1])) return sum; 104 if (out(x1, y1, x2, y2, mn[0], mn[1], mx[0], mx[1])) return 0; 105 if (in(x1, y1, x2, y2, p[0], p[1], p[0], p[1])) res += p.v; 106 if (son[0]) res += son[0] -> query(x1, y1, x2, y2); 107 if (son[1]) res += son[1] -> query(x1, y1, x2, y2); 108 return res; 109 } 110 } *T; 111 112 void work_add() { 113 static point p; 114 p.get(); 115 if (!T) T = new(p)KD_tree; 116 else T -> insert(p, 0); 117 } 118 119 inline void work_query() { 120 static int x1, y1, x2, y2; 121 x1 = read(), y1 = read(), x2 = read(), y2 = read(); 122 if (!T) printf("%lld ", ans = 0); 123 else printf("%lld ", ans = T -> query(x1, y1, x2, y2)); 124 } 125 126 int main() { 127 int i, oper; 128 n = read(0), ans = 0; 129 while (1) { 130 oper = read(0); 131 if (oper == 1) work_add(); 132 else if (oper == 2) work_query(); 133 else break; 134 } 135 return 0; 136 }