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题目链接：

id=3067

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

13 4 41 42 33 23 1

Sample Output

Test case 1: 5

Source

题意：日本岛东海岸与西海岸分别有N和M个城市，如今修快速公路连接东西海岸的城市，求交点个数。

做法：（做法的解释来自：）记每条告诉公路为(x,y), 即东岸的第x个城市与西岸的第y个城市修一条路。当两条路有交点时，满足（x1-x2）*（y1-y2） < 0。所以，将每条路按x从小到达排序，若x同样，按y从小到大排序。 然后按排序后的公路用树状数组在线更新。求y的逆序数之 和 即为交点个数。

上面说的可能有点难理解，具体说明例如以下。

记第i条边的端点分别为xi,yi。

由于x是从小到大排序的，如果当前我们在处理第k条边，那么第1~k-1条边的x必定是小于（等于时候暂且不讨论）第k条边的 x 的。那么前k-1条边中，与第k条边相交的边的y值必定大于yk的，所以此时我们仅仅须要求出在前k-1条边中有多少条边的y值在区间[yk, M]就可以,也就是求yk的逆序数，M为西岸城市个数，即y的最大值。

所以就将问题转化成区间求和的问题，树状数组解决。当两条边的x同样时，我们记这两条边的y值分别为ya，yb（ya<yb),我们先处理（x，ya），再处理（x，yb），原因非常明显，由于当x同样时。这两条边是觉得没有交点的，若先处理（x，yb）。那么下次处理（x。ya）时，（x，ya）就会给（x，yb）添加一个逆序，也就是将这两条边做相交处理了。

代码例如以下：

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;int n,m,k;#define MAX 2047struct node{	int l,r;//分别为左端点和右端点}line[MAX*MAX];//快速公路的结构int c[MAX];//树状数组bool cmp( node a, node b){	if(a.l==b.l)	{		return a.r
         
           0)	{		sum += c[x];		x -= Lowbit(x);	}	return sum;}int main(){	int t, i;	scanf("%d",&t);	int tt=1;	while(t--)	{		scanf("%d %d %d",&n,&m,&k);		for(i = 1; i <= k; i++) //i须从1開始		{			scanf("%d%d",&line[i].l,&line[i].r);//输入		}		sort(line+1,line+k+1,cmp);//依照l的从小到大排序，l同样时按r的从小到大排序，		//这样就形成了r的一维树状数组		memset(c,0,sizeof(c));		__int64 ret=0;//最后结果		for(i = 1; i <= k; i++) //i须从1開始		{			update(line[i].r,1);//插入树状数组中			ret+=i-sum(line[i].r);//i为当前已插入的元素的个数。sum返回了小于等于当前r值的元素个数，			//相减即为满足条件的元素个数		}		printf("Test case %d: %lld\n",tt++,ret);	}	return 0;}