poj3067 Japan

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23937		Accepted: 6473

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

Source

Southeastern Europe 2006

给定k条线段，求交点个数(不包括端点)

这是一道树状数组裸题…不过自己提交了15次才过了...QAQ

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i.=n;i--)
#define MAXN 2005
#define LL long long
#define pa pair<int,int>
using namespace std;
struct data
{
	int x,y;
}a[MAXN*MAXN];
int t,n,m,k,f[MAXN];
LL sum;
int read()
{
	int ret=0,flag=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	return ret*flag;	
}
bool cmp(data d1,data d2)
{
	if (d1.x!=d2.x) return d1.x>d2.x;
	else return d1.y>d2.y;
}
void add(int k)
{
	for(int i=k;i<=m;i+=i&(-i)) f[i]++;
}
LL getsum(int k)
{
	LL ret=0;
	for(int i=k;i>0;i-=i&(-i)) ret+=f[i];
	return ret;
}
int main()
{
	t=read();
	F(ii,1,t)
	{
		memset(f,0,sizeof(f));
		n=read();m=read();k=read();
		sum=0;
		F(i,1,k) a[i].x=read(),a[i].y=read();
		sort(a+1,a+k+1,cmp);
		F(i,1,k)
		{
			sum+=getsum(a[i].y-1);
			add(a[i].y);
		}
		printf("Test case %d: %lld\n",ii,sum);
	}
}