【LG3249】[HNOI2016]矿区

【LG3249】[HNOI2016]矿区

题面

洛谷

题解

先平面图转对偶图，

建好了对偶图之后随意拿出一个生成树，以无边界的范围为根。

无边界的范围很好求，用叉积算出有向面积时，算出来是负数的就是无边界的范围。

然后标记所有的树边，记录生成树中每个子树的矿区面积和及面积平方和。

对于每一个询问，先找到询问里出现的边，如果有非树边就忽略，否则如果这条边所在的面是儿子，就加上子树的面积，如果是父亲就减去儿子子树的面积。

这个可以通过画图手玩进行证明。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
#include <vector> 
using namespace std; 
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 2e5 + 5, MAX_M = 1.2e6 + 5; 
const double EPS = 1e-10; 
struct Point { int x, y; } p[MAX_N]; 
Point operator - (const Point &l, const Point &r) { return (Point){l.x - r.x, l.y - r.y}; } 
long long cross(const Point &l, const Point &r) { return 1ll * l.x * r.y - 1ll * l.y * r.x; } 
struct Edge { int id, x, y; double a; } e[MAX_M]; int e_cnt = 0; 
bool operator < (const Edge &l, const Edge &r)
    { return fabs(l.a - r.a) < EPS ? l.y < r.y : l.a < r.a; } 
vector<Edge> G[MAX_N], T[MAX_M]; 
void Add_Edge(int u, int v) { 
	e[e_cnt] = (Edge){e_cnt, u, v, atan2(p[v].y - p[u].y, p[v].x - p[u].x)}; 
	G[u].push_back(e[e_cnt]); e_cnt++; 
} 
int N, M, Q; 
int cnt, root, nxt[MAX_M], pos[MAX_M], fa[MAX_M];
long long ans1, ans2, s1[MAX_M], s2[MAX_M]; 
void build() { 
	for (int i = 1; i <= N; i++) sort(G[i].begin(), G[i].end()); 
	for (int i = 0; i < e_cnt; i++) { 
		int y = e[i].y;
		vector<Edge> :: iterator ite = lower_bound(G[y].begin(), G[y].end(), e[i ^ 1]); 
		if (ite == G[y].begin()) ite = G[y].end(); 
		--ite, nxt[i] = ite -> id; 
	} 
	for (int i = 0; i < e_cnt; i++) { 
		if (pos[i]) continue; 
		pos[i] = pos[nxt[i]] = ++cnt;
		for (int j = nxt[i]; e[j].y != e[i].x; j = nxt[j], pos[j] = cnt)
			s1[cnt] += cross(p[e[j].x] - p[e[i].x], p[e[j].y] - p[e[i].x]); 
		if (s1[cnt] <= 0) root = cnt; 
	} 
	for (int i = 0; i < e_cnt; i++)
		T[pos[i]].push_back((Edge){i, pos[i], pos[i ^ 1]}); 
}
bool vis[MAX_M], ist[MAX_M]; 
void dfs(int x) { 
	s2[x] = 1ll * s1[x] * s1[x], s1[x] <<= 1ll, vis[x] = 1; 
	for (int i = 0, sz = T[x].size(); i < sz; i++) { 
		int y = T[x][i].y; if (vis[y]) continue; 
		ist[T[x][i].id] = ist[T[x][i].id ^ 1] = 1; 
		fa[y] = x; dfs(y); s1[x] += s1[y], s2[x] += s2[y]; 
	} 
} 
long long gcd(long long x, long long y) { 
	if (y == 0) return x; 
	else return gcd(y, x % y); 
} 
int q[MAX_M]; 
int main () {
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin);
#endif
	N = gi(), M = gi(), Q = gi(); 
	for (int i = 1; i <= N; i++) p[i] = (Point){gi(), gi()}; 
	for (int i = 1; i <= M; i++) {
		int u = gi(), v = gi(); 
		Add_Edge(u, v), Add_Edge(v, u); 
	} 
	build(); dfs(root); 
	while (Q--) {
		int n = (gi() + ans1) % N + 1; 
		for (int i = 1; i <= n; i++) q[i] = (gi() + ans1) % N + 1; 
		q[n + 1] = q[1], ans1 = ans2 = 0; 
		for (int i = 1; i <= n; i++) { 
			int x = q[i], y = q[i + 1]; 
			Edge e = (Edge){0, x, y, atan2(p[y].y - p[x].y, p[y].x - p[x].x)}; 
			vector<Edge> :: iterator ite = lower_bound(G[x].begin(), G[x].end(), e); 
			int j = ite -> id; if (!ist[j]) continue; 
			if (fa[pos[j]] == pos[j ^ 1]) ans1 += s2[pos[j]], ans2 += s1[pos[j]]; 
			else ans1 -= s2[pos[j ^ 1]], ans2 -= s1[pos[j ^ 1]]; 
		} 
		long long tmp = gcd(ans1, ans2); 
		ans1 /= tmp, ans2 /= tmp; 
		printf("%lld %lld
", ans1, ans2); 
	} 
    return 0; 
}