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.

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.

For each test case write one line on the standard output: 
Test case (case number): (number of crossings)

1

3 4 4

1 4

2 3

3 2

3 1

Test case 1: 5

#include<cstdio>

#include<algorithm>

#define MAX 1000*1000+3

using namespace std;

struct Road{

    int u,v;

}road[MAX],tmp[MAX];

int n,m,k;

long long cnt;

bool cmp(Road a,Road b)

{

    if(a.u==b.u) return a.v<b.v;

    else return a.u<b.u;

}

void Merge(int l,int m,int r)

{

    int i = l;

    int j = m + ;

    int k = l;

    while(i <= m && j <= r)

    {

        if(road[i].v > road[j].v)

        {

            tmp[k++] = road[j++];

            cnt += m - i + ;

        }

        else

        {

            tmp[k++] = road[i++];

        }

    }

    while(i <= m) tmp[k++] = road[i++];

    while(j <= r) tmp[k++] = road[j++];

    for(int i=l;i<=r;i++) road[i] = tmp[i];

}

void Merge_sort(int l,int r)

{

    if(l < r)

    {

        int m = (l + r) >> ;

        Merge_sort(l,m);

        Merge_sort(m+,r);

        Merge(l,m,r);

    }

}

int main()

{

    int t;

    scanf("%d",&t);

    for(int kase=;kase<=t;kase++)

    {

        scanf("%d%d%d",&n,&m,&k);

        for(int i=;i<=k;i++) scanf("%d%d",&road[i].u,&road[i].v);

        sort(road+,road+k+,cmp);

        cnt=;

        Merge_sort(,k);

        printf("Test case %d: %I64d\n",kase,cnt);

    }

}

 #include<cstdio>

 #include<algorithm>

 #define MAXN 1000*1000+3

 using namespace std;

 int n,m,k;

 struct Road{

     int u,v;

 }road[MAXN];

 bool cmp(Road a,Road b)

 {

     if(a.u==b.u) return a.v<b.v;

     else return a.u<b.u;

 }

 //BIT - 单点增加，区间查询 - st

 struct _BIT{

     int N,C[MAXN];

     int lowbit(int x){return x&(-x);}

     void init(int n)//初始化共有n个点

     {

         N=n;

         for(int i=;i<=N;i++) C[i]=;

     }

     void add(int pos,int val)//在pos点加上val

     {

         while(pos<=N)

         {

             C[pos]+=val;

             pos+=lowbit(pos);

         }

     }

     int sum(int pos)//查询1~pos点的和

     {

         int ret=;

         while(pos>)

         {

             ret+=C[pos];

             pos-=lowbit(pos);

         }

         return ret;

     }

 }BIT;

 //BIT - 单点增加，区间查询 - ed

 int main()

 {

     int t;

     scanf("%d",&t);

     for(int kase=;kase<=t;kase++)

     {

         scanf("%d%d%d",&n,&m,&k);

         for(int i=;i<=k;i++) scanf("%d%d",&road[i].u,&road[i].v);

         sort(road+,road+k+,cmp);

         long long cnt=;

         BIT.init(m);

         for(int i=;i<=k;i++)

         {

             BIT.add(road[i].v,);//road[i].v出现一次，加上

             cnt+=BIT.sum(m)-BIT.sum(road[i].v);//计算出在目前情况下，有多少个比road[i].v大的数出现过

         }

         printf("Test case %d: %I64d\n",kase,cnt);

     }

 }