POJ 3067

求相交的公路，对N排序，然后对M求逆序对，不过要求有相交的，必然要求在次城市后面，所以int sumge = i - (qur(s[tmp].y)) - ge;，这里修改了老多次，不过还好1Y了。

#define lowbit(x) x & -x
struct sta
{
    int x,y;
};
sta s[MAXN];
int n,m,k,maxx;
int num[1005];
int cmp(const void *a,const void *b)
{
    struct sta *c = (sta *)a;
    struct sta *d = (sta *)b;
    if(c->x == d->x)
    {
        return c->y - d->y;
    }
    return c->x - d->x;
}
void add(int x,int va)
{
    while(x <= m)
    {
        num[x] += va;
        x += lowbit(x);
    }
}
int qur(int x)
{
    int s=0;
    while(x)
    {
        s += num[x];
        x -= lowbit(x);
    }
    return s;
}
int main()
{
    //FOPEN;
    int ces,a,b,kk = 0;SCF(ces);
    while(ces--)
    {
        SET(num,0);
        int tmp;
        long long sum=0;
        SCFT(n,m,k);
        F(i,k)
        {
            SCFD(a,b);
            s[i].x=a;
            s[i].y=b;
        }
        qsort(s,k,sizeof(s[0]),cmp);
        //DB(s[3].y);
        for(int i=0;i<k;)
        {
            tmp = i;
            while((i<k)&&s[i].y==s[tmp].y)
                i++;
            int ge = i - tmp;
            int sumge = i - (qur(s[tmp].y)) - ge;
            sum += sumge*ge;
            //DB("ge="<<ge<<"  sumge="<<sumge<<"  qur="<<qur(s[tmp].y));
            //DB(sum<<"--");
            add(s[tmp].y,ge);
        }
        printf("Test case %d: %lld\n",++kk,sum);
    }
}