Poj2481-tree Array application

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For this problem, which has two keywords and is monotonic, the general method for tree array optimization is to fix a keyword (or the endpoint of the interval ), then, query the first n items and Properties of another keyword to determine the answer to the current object.

1. first sort by S from small to big, if S is the same according to T from big to small sort (so that the previous impact after) use a tree array to query the number of more current than the E-1, pay special attention to handling the exact same interval.
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2. The plug-in Q & amp; A template is the number of smaller numbers before this number.
For this question, t is first sorted in descending order. If they are the same, S is sorted in ascending order, and then the number of s smaller than the current value is sum (I ): that is, the number of smaller than I, and then modify (I );

Poj 2481 cows solution: similar to stars,
FJ has a nheaded ox (numbered 1 ~ N), each cow has a test value [s, E, when their test values meet the following conditions, it proves that Niu I is stronger than Niu J: Si <= SJ and j <= EI and EI-Si> ej-SJ. It is now known that each cow's test value is required to output the number of cows per cow to be stronger than its own.

Package tree array _ insert question segment ;//!!!!!!!!!! // When a property in the question is monotonically decreasing, I-sum (I-1) can be used to query a number larger than I in a tree array; import Java. io. bufferedreader; import Java. io. ioexception; import Java. io. inputstreamreader; import Java. io. outputstreamwriter; import Java. io. printwriter; import Java. io. streamtokenizer; import Java. util. arrays; public class poj2481two {static int max = 100010; static int C [] = new int [Max]; static int ans [] = new int [Max]; static d [] D; static streamtokenizer in = New streamtokenizer (New bufferedreader (New inputstreamreader (system. in); static printwriter out = new printwriter (New outputstreamwriter (system. out); Final Static int nextint () throws ioexception {In. nexttoken (); Return (INT) in. nval;} static int lowbit (int I) {int ans = I & (-I); Return ans;} // Insert the vertex question segment ,,,,,,,,, this template is used to query the number of smaller numbers before it. Static void modify (int K, int d) {While (k <= max) {C [k] + = D; K + = lowbit (k) ;}} static int Sum (int n) {int result = 0; while (n> 0) {result + = C [N]; n-= lowbit (n) ;}return result ;} public static void main (string [] ARGs) throws ioexception {int num; while (num = nextint ())! = 0) {arrays. fill (C, 0); arrays. fill (ANS, 0); D = new D [num]; for (INT I = 0; I <num; I ++) {d [I] = new D (nextint () + 1, nextint () + 1); // cannot start from scratch d [I]. index = I;} arrays. sort (D, 0, num); For (INT I = 0; I <num; I ++) {if (I> 0 & D [I]. S = d [I-1]. S & D [I]. E = d [I-1]. e) {// modify (COW [I]. s, 1); ans [d [I]. index] = ans [d [I-1]. index];} else {ans [d [I]. index] = I-sum (d [I]. e-1);} modify (d [I]. e, 1);} For (INT I = 0; I <num-1; I ++) out. print (ANS [I] + ""); out. println (ANS [num-1]);} Out. flush (); out. close () ;}} Class D implements comparable <D> {int s; int e; int index; Public D (int s, int e) {This. S = s; this. E = e ;}@ overridepublic int compareto (d o) {// todo auto-generated method stubif (this. S <O. s) Return-1; else if (this. S = O. s) {If (this. e> O. e) Return-1; elsereturn 1;} elsereturn 1 ;}}