Poj 2761 feed the dogs calculates the number of K in the range

The key point of this question is that you can use a tree array: All query intervals do not contain each other, but may overlap.

In this way, you can add and delete edges from left to right using a tree array.

If there is an inclusion relationship, you cannot use a tree array. For the reason, see the sample data of poj 2104.

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# Include <cstdio> # Include <Cstring> # Include <Algorithm> Using   Namespace  STD;  Const   Int Maxn = 100010  ;  Struct Node {  Int  Key, ID;  Bool   Operator <( Const Node & cm) Const  {  Return Key < Cm. Key ;}} dog [maxn];  Struct  PP {  Int  L, R, kx, ID;  Bool  Operator <( Const PP & cm) Const  {  If (L! = Cm. l) Return L < Cm. L;  Return R < Cm. r ;}} Q [maxn];  Int  C [maxn], X [maxn], ran [maxn], ANS [maxn];  Void Update ( Int X, Int D ){  For (; X <maxn; x + = x &- X) C [x] + = D ;}  Int Find_kth ( Int  K ){  Int Ans = 0 , CNT = 0  , I;  For (I = 20 ; I> = 0 ; I --) {Ans + = ( 1 < I );  If (ANS> = maxn | CNT + C [ANS]> = K) ans -= ( 1 < I );  Else  CNT + = C [ANS];}  Return Ans + 1  ;}  Int Main (){  Int  N, m, I, J, K, a, B; scanf (  "  % D  " , & N ,& M );  For (I = 1 ; I <= N; I ++ ) {Scanf (  "  % D  " ,& DOG [I]. Key); dog [I]. ID = I;} Sort (dog + 1 , Dog + 1 + N); X [ 1 ] = Dog [ 1 ]. Key; ran [DOG [ 1 ]. ID] = 1  ;  For (J = 1 , I = 2 ; I <= N; I ++ ){  If (DOG [I]. Key! = Dog [I- 1 ]. Key) x [++ J] =DOG [I]. Key; ran [DOG [I]. ID] = J ;}  For (I = 1 ; I <= m; I ++ ) {Scanf (  "  % D  " , & A, & B ,& K );  If (A> B) Swap (a, B); Q [I]. L = A; Q [I]. r = B; Q [I]. kx = K; Q [I]. ID = I;} Q [  0 ]. L =1 ; Q [ 0 ]. R = 0  ; Sort (Q + 1 , Q + M + 1  );  For (I = 1 ; I <= m; I ++ ){  For (J = Q [I- 1 ]. L; j <= Q [I- 1 ]. R & J <q [I]. L; j ++ ) Update (ran [J], - 1  );  For (J = Q [I]. L <q [I- 1 ]. R + 1 ? Q [I- 1 ]. R + 1 : Q [I]. L; j <= Q [I]. R; j ++ ) Update (ran [J],  1  );  Int Num = Find_kth (Q [I]. kx); ans [Q [I]. ID] = X [num];} For (I = 1 ; I <= m; I ++) printf ( "  % D \ n  "  , ANS [I]);  Return   0  ;}