HDU 4970 killing monster from generation to generation brush QQ do not need line tree ~

[Question] Tower anti-DDoS game, with N-long roads, has m gunshots, which can cover the range of [Li, Ri] and power ci, that is, every second, the turret can cause CI point damage to monsters in the range. There are only Q monsters, each of which has hi-point blood, and the position is Xi. When the monster's blood volume is reduced to 0 or less, the monster keeps moving toward n. Ask how many monsters can leave this path.

[Problem solution] We can use a line segment tree, but fortunately we have a QQ brush algorithm from generation to generation, making the solution much simpler ~ Pai_^

 1 # Include <iostream> 2 # Include <cstdio> 3 # Include <cstring> 4 # Include <cmath> 5 # Include <algorithm> 6   7  Using   Namespace  STD;  8 _ Int64 num [ 100005  ];  9   10   Struct Node {__ int64 HP, X;} m [ 100005  ];  11   12   Int  CMP (node A, Node B)  13  {  14       Return A. x < B. X;  15   }  16   17   Int  Main ()  18   {  19       Int  N, K, a, B, d;  20       While (~ Scanf ("  % D  " ,& N), n)  21   {  22 Scanf ( "  % D  " ,& K );  23 Memset (Num, 0 , Sizeof  (Num ));  24          While (K -- )  25   {  26 Scanf ( "  % D  " , & A, & B ,& D );  27 Num [a] + = D;  28 Num [B + 1 ]-= D;  29  }  30 _ Int64 sum = 0  ;  31           For ( Int I = 0 ; I <= N; I ++ )  32   {  33 Sum + = Num [I];  34 Num [I] = SUM; 35   }  36           // Num [I]:  Attack Strength at each point I  37 Scanf ( "  % D  " ,& K );  38           For ( Int I = 0 ; I <K; I ++ )  39 Scanf ( "  % I64d % i64d  " , & M [I]. HP ,& M [I]. X );  40 Sort (M, M + K, CMP );  41 Sum = 0  ;  42           Int Ans = 0 , J = k- 1  ;  43          For ( Int I = N; I> = 0 ; I -- )  44   {  45 Sum + = num [I]; //  Sum: The total attack strength from point N to point I  46               While (I = m [J]. X) //  Several Monsters may have emerged from point I.  47   { 48                   If (M [J]. HP> Sum)  49 Ans ++ ;  50 J -- ;  51                   If (J < 0  )  52                       Break  ;  53   } 54               If (J < 0  )  55                   Break  ;  56   }  57 Printf ( "  % D \ n  "  , ANS );  58   }  59      Return   0  ;  60 }