POJ 3067 Japan (tree array/reverse order number)

Japan Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22258 Accepted: 5995 Description Japan plans to welcome the ACM ICPC World Finals and a lot of roads must is built for the venue. Japan is Tall island with N cities on the East coast and M cities on the West coast (M <=, N <= 1000). K superhighways would be build. Cities on each coast is numbered 1, 2, ... Each superhighway are straight line and connects city in the East coast with city of the West coast. The funding for the construction are guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At the most of the superhighways cross at the one location. Write A program that calculates the number of the crossings between superhighways. Input The input file starts with t-the number of test cases. Each test case is starts with three numbers–n, M, K. Each of the next K lines contains, numbers–the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one are the number of the city of the West coast. Output For each test case, write one line in the standard output: Test Case Number: (Number of crossings) Sample Input 13 4 41 42 33 23 1 Sample Output Test Case 1:5 Source Southeastern Europe 2006 #include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include < Stdlib.h>using namespace Std;int n,m,k;__int64 c[1000100];struct node{int x; int y;} Q[1000500];bool CMP (node A,node b) {if (a.x==b.x) {return a.y<=b.y; } else {return a.x<b.x; }}int lowbit (int x) {return x& (-X);} void Updata (int i,int t) {while (i<=m) {C[i] + = t; i = i + lowbit (i); }}__int64 getsum (int i) {__int64 sum = 0; while (i>0) {sum + = C[i]; i = I-lowbit (i); } return sum;} int main () {int T; int p = 0; scanf ("%d", &t); while (t--) {scanf ("%d%d%d", &n,&m,&k); Memset (C,0,sizeof (c)); for (int i=0; i<k; i++) {scanf ("%d%d", &q[i].x,&q[i].y); } __int64 ans = 0; Sort (q,q+k,cmp); for (int i=0; i<k; i++) {updata (q[i].y,1); Ans + = GetSUM (m)-getsum (Q[I].Y); } printf ("Test Case%d:%i64d\n", ++p,ans); } return 0;}

POJ 3067 Japan (tree array/reverse order number)