51nod1107(斜率小于0的连线个数)

题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1107

先分别以 x 和 y 为第一和第二关键字从小到大排序，然后对 y 求逆序对……

一开始竟然想用线段树来做……

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;const LL maxn = 50000 + 10;struct Point{    LL x, y;    Point() {}    Point(LL x, LL y):x(x), y(y) {}}P[maxn];LL a[maxn];LL r[maxn];LL n;LL ans;int cmp(const Point& a, const Point& b){    if (a.x == b.x){        return a.y < b.y;    }    return a.x < b.x;}void msort(LL s, LL t){    LL m, i, j, k;    if (s == t) return ;    m = (s + t) >> 1;    msort(s, m);    msort(m+1, t);    i = s;    j = m + 1;    k = s;    while (i <= m && j <= t){        if (a[i] > a[j]){            r[k++] = a[i++];            ans += (LL)(t - j + 1);        }else{            r[k++] = a[j++];        }    }    while (i <= m){        r[k++] = a[i++];    }    while (j <= t){        r[k++] = a[j++];    }    for (i=s; i<=t; i++) a[i] = r[i];}int main(){    scanf("%lld", &n);    for (LL i=1; i<=n; i++) scanf("%lld%lld", &P[i].x, &P[i].y);    sort(P+1, P+n+1, cmp);    for (LL i=1; i<=n; i++) a[i] = P[i].y;    msort(1, n);    cout << ans << endl;    return 0;}