Japan POJ - 3067 转化思维 转化为求逆序对

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 OutputTest case 1:

题目大意：一个平面，左边自上而下排列了N个点，
标号为1,...,N，
右边自上而下排列了M个点，
标号为1,...,M，它们之间有K条线段相连
，每条线段有两个值：x,y，
表示该线段连接了左边的标号为x的点和右边的标号为y的点，
问有多少个交点
仔细想下  这不就是求逆序对吗

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug         printf("******\n")
#define mem(a,b)    memset(a,b,sizeof(a))
#define fuck(x)     cout<<"["<<x<<"]"<<endl
#define f(a)        a*a
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define lowbit(x)   x&-x
#pragma comment (linker,"/STACK:102400000,102400000")

using namespace std;
typedef long long LL ;
const int maxn = 1e6 + ;
LL c[maxn];
int t, n, m, k, cas = ;
struct node {
    int x, y;
} a[maxn];
int cmp(node a, node b) {
    if (a.x == b.x) return a.y < b.y;
    return a.x < b.x;
}
void update(int x) {
    while(x <= m) {
        c[x] += ;
        x += lowbit(x);
    }
}
int  getans(int x) {
    int s = ;
    for(int i = x; i > ; i -= lowbit(i))
        s += c[i];
    return s;
}
int main() {
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &n, &m, &k);
        memset(c, , sizeof(c));
        for (int i =  ; i <= k ; i++)
            scanf("%d%d", &a[i].x, &a[i].y);
        sort(a + , a + k + , cmp);
        LL ans = ;
        for (int i =  ; i <= k ; i++) {
            ans += getans(m) - getans(a[i].y);
            update(a[i].y);
        }
        printf("Test case %d: %lld\n", cas++, ans);
    }
    return ;
}