HDU - 3465 Life is a Line

题意：求在[L,R]范围内直线交点的对数

思路：首先如果两条直线相交的话，那么(xl1-xl2)*(yl1-yl2)<0，这刚好满足求逆序数对的条件

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;const int MAXN = 50010;struct node{double x,y;int id;}a[MAXN];int n,m,cnt;double L,R;int c[MAXN];int lowbit(int x){return x&(-x);}bool cmp1(node a,node b){return a.x < b.x;}bool cmp2(node a,node b){return a.y < b.y;}void update(int x,int d){while (x < MAXN){c[x] += d;x += lowbit(x);}}int sum(int x){int ans = 0;while (x > 0){ans += c[x];x -= lowbit(x);}return ans;}int main(){while (scanf("%d",&n) != EOF){m = cnt = 0;scanf("%lf%lf",&L,&R);for (int i = 1; i <= n; i++){double x1,x2,y1,y2;scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);if (x1 == x2){if (x1 > L && x1 < R)m++;}else {cnt++;a[cnt].x = y1+(L-x1)*(y2-y1)/(x2-x1);a[cnt].y = y1+(R-x1)*(y2-y1)/(x2-x1);}}n = cnt;sort(a+1,a+1+n,cmp1);a[0].x = a[1].x - 1;for (int j = 0,i = 1; i <= n; i++)if (a[i].x == a[i-1].x)a[i].id = j;else a[i].id = ++j;sort(a+1,a+1+n,cmp2);memset(c,0,sizeof(c));ll ans = 0;for (int i = 1; i <= n; i++){ans += i-1-sum(a[i].id);update(a[i].id,1);}ans += (ll)m*n;cout << ans << endl;}return 0;}