Geometers Anonymous Club CodeForces - 1195F (闵可夫斯基和)

大意: 给定$n$个凸多边形, $q$个询问, 求$[l,r]$内闵可夫斯基区间和的顶点数.

要用到一个结论, 闵可夫斯基和凸包上的点等于向量种类数.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <map>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
using namespace std;
typedef pair<int,int> pii;
const int N = 1e6+10;
int n, q, cnt, x[N], y[N], c[N];
int L[N], R[N], ans[N];
struct _ {int l,r;} b[N];
vector<int> g[N];
pii a[N];
map<pii,int> pre;

int gcd(int a,int b) {return b?gcd(b,a%b):a;}
pii reduce(int x, int y) {
	int g = gcd(abs(x),abs(y));
	if (g) x/=g,y/=g;
	return pii(x,y);
}
void add(int x, int v) {
	for (; x<=cnt; x+=x&-x) c[x]+=v;
}
int query(int x) {
	int r = 0;
	for (; x; x^=x&-x) r+=c[x];
	return r;
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) {
		int k;
		scanf("%d", &k);
		REP(j,0,k-1) scanf("%d%d",x+j,y+j);
		L[i] = cnt+1;
		REP(j,0,k-1) a[++cnt] = reduce(x[j]-x[(j+1)%k],y[j]-y[(j+1)%k]);
		R[i] = cnt;
	}
	scanf("%d", &q);
	REP(i,1,q) {
		int l, r;
		scanf("%d%d", &l, &r);
		b[i].l = L[l];
		b[i].r = R[r];
		g[b[i].r].pb(i);
	}
	int now = 1;
	REP(i,1,cnt) {
		if (pre[a[i]]) add(pre[a[i]],-1);
		pre[a[i]] = i;
		add(i,1);
		for (int j:g[i]) ans[j] = query(b[j].r)-query(b[j].l-1);
	}
	REP(i,1,q) printf("%d\n", ans[i]);
}