POJ 3067 Japan (树状数组求逆序对)

题意：

首先一行是T组数据，每组数据第一行是N,M,K，N代表左侧点的数量，M代表右侧点的数量，K代表有K条连线

次下K行有两个值X，Y代表左边第X和右边第Y个点相连。

问这些连线最多有几个交点。（同起点的线不相交）

思路：

显然，无论插入线段的顺序是怎么样的，交点的个数都不会改变

根据左边从小到大排序，求右边的逆序数就行

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=1000000;
typedef struct Node{
int x;
int y;
bool operator < (const Node &a)const{
if(a.x==x)
return a.y>y;
return a.x>x;
}
}Node;
int b[MAXN];
Node a[MAXN];
int lowbit(int x){
return x&(-x);
}
void modify(int x,int add){
while(x<=MAXN){
b[x]+=add;
x+=lowbit(x);
}
}
int get_sum(int x){
int ret=0;
while(x!=0){
ret+=b[x];
x-=lowbit(x);
}
return ret;
}
int main()
{
//ios::sync_with_stdio(false);
int T;
scanf("%d",&T);
for(int ppp=1;ppp<=T;ppp++){
int x,y,k;
scanf("%d%d%d",&x,&y,&k);
int n=max(x,y);
long long ans=0;
memset(b,0,sizeof(b));
for(int i=0;i<k;i++)
scanf("%d%d",&a[i].x,&a[i].y);
sort(a,a+k);
for(int i=0;i<k;i++){
modify(a[i].y,1);
ans+=get_sum(n)-get_sum(a[i].y);
}
printf("Test case %d: %lld\n",ppp,ans);
}
}

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