POJ 3067 Japan （樹狀數組）

POJ 3067 Japan （樹狀數組）

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at 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 starts with three numbers – N, M, K. Each of the next K lines contains two 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 is the number of the city of the West coast.

Output

For each test case write one line on the standard output:
Test case (case number): (number of crossings)

Sample Input

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5

這題每個點可以發出多條邊，只需按照每條邊的端點大小排序，先按照左邊從大到小，然後右邊從大到小。 這樣處理之後，左邊的端點依次標記為1 ~ num。不會對結果造成影響。然後就是赤裸裸的逆序對的問題了。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 1000000000;
const int maxn = 100005;
int t, n, m, k, f[1000005];
struct C {
    int st, en;
} in[1000005];
LL c[1000005];
bool cmp(C x, C y) {
    if(x.st != y.st) return x.st < y.st;
    return x.en < y.en;
}
int main()
{
   // freopen("in.txt", "r", stdin);
    cin >> t;
    for(int ca = 1; ca <= t; ca++) {
        cin >> n >> m >> k;
        for(int i = 0; i < k; i++) scanf("%d%d", &in[i].st, &in[i].en);
        sort(in, in+k, cmp);
        int num = 0;
        for(int i = 0; i < k; i++) {
            num++;
            in[i].st = num;
            f[num] = in[i].en;
        }
        memset(c, 0, sizeof(c));
        LL sum = 0;
        for(int i = 1; i <= num; i++) {
            for(int j = f[i]+1; j <= maxn; j += j&-j) sum += c[j];
            for(int j = f[i]; j > 0; j -= j&-j) c[j]++;
        }
        printf("Test case %d: %I64d\n", ca, sum);
    }

    return 0;
}