poj3067 Japan

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
这题和star,cows差不多，把读入的坐标看做x,y，那么先按x进行升序排序，如果x相同的话就按照y升序排序。然后依次循环，加上比a[i].y大的线段的个数。注意用__int64

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 3005
int b[maxn];
struct node{
	int x,y;
}a[maxn*maxn];

bool cmp(node a,node b){
	if(a.x==b.x)return a.y<b.y;
	return a.x<b.x;
}

int lowbit(int x){
	return x&(-x);
}
void update(int pos,int num)
{
	while(pos<=maxn){
		b[pos]+=num;pos+=lowbit(pos);
	}
}
int getsum(int pos)
{
	int num=0;
	while(pos>0){
		num+=b[pos];pos-=lowbit(pos);
	}
	return num;
}

int main()
{
	int n,m,i,j,T,k,num1=0;
	__int64 sum;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&n,&m,&k);
		for(i=1;i<=k;i++){
			scanf("%d%d",&a[i].x,&a[i].y);
			a[i].x++;a[i].y++;
		}
		sort(a+1,a+1+k,cmp);
		memset(b,0,sizeof(b));
		sum=0;
		for(i=1;i<=k;i++){
			sum+=i-1-getsum(a[i].y);
			update(a[i].y,1);
		}
		num1++;
		printf("Test case %d: %I64d\n",num1,sum);
	}
	return 0;
}