Japan
Time Limit: 1000MS |
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Memory Limit: 65536K |
Total Submissions: 25489 |
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Accepted: 6907 |
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 Parse: Tree-like Array (single point update, interval query). The road is sorted (by W from small to large sort, w is the same as e from small to large sort), so, for road ri, if Ej>ei (0≤j<i), RI, RJ Two roads there are intersections. With a tree array, we can see how many roads in front of the road ri are less than or equal to EI, and then subtract it to get the number of the e greater than EI for the road ahead of the RI, and the sum is the result.
#include <cstdio> #include <cstring> #include <algorithm> #define LOWBIT (x) (x) & (-X) #define LL Long longusing namespace Std;struct road{int W, E; BOOL operator < (const road& b) Const {if (w! = B.W) return w<b.w; Return e<b.e; }}r[1000005];int c[1005];int N, M, k;void Add (int x, int val) {for (int i = x; I <=, i + = Lowbit (i)) c[i] + = val;} int sum (int x) {int ret = 0; for (int i = x; i > 0; I-= Lowbit (i)) ret + = C[i]; return ret;} void Solve () {memset (c, 0, sizeof (c)); Sort (R, R+k); ll res = 0; for (int i = 0; i < K; ++i) {res + = I-sum (R[I].E); Add (r[i].e, 1); } printf ("%i64d\n", res);} int main () {int T, CN = 0; scanf ("%d", &t); while (t--) {scanf ("%d%d%d", &n, &m, &k); for (int i = 0; i < K; ++i) scanf ("%d%d", &R[I].W, &R[I].E); printf ("Test Case%d:", ++CN); Solve (); } return 0;}
POJ 3067 Japan