BZOJ2564: 集合的面积(闵可夫斯基和 凸包)

题意

题目链接

Sol

这个东西的学名应该叫“闵可夫斯基和”。就是合并两个凸包

首先我们先分别求出给出的两个多边形的凸包。合并的时候直接拿个双指针扫一下，每次选最凸的点就行了。

复杂度(O(nlogn + n))

#include<bits/stdc++.h>
#define LL long long 
//#define int long long 
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, M;
struct Point {
	LL x, y;
	Point operator - (const Point &rhs) const {
		return {x - rhs.x, y - rhs.y};
	}
	Point operator + (const Point &rhs) const {
		return {x + rhs.x, y + rhs.y};
	}
	LL operator ^ (const Point &rhs) const {
		return x * rhs.y - y * rhs.x;
	}
	bool operator < (const Point &rhs) const {
		return x == rhs.x ? y < rhs.y : x < rhs.x;
	}
	bool operator == (const Point &rhs) const {
		return x == rhs.x && y == rhs.y;
	}
	bool operator != (const Point &rhs) const {
		return x != rhs.x || y != rhs.y;
	}
};
vector<Point> v1, v2;
Point q[MAXN];
int top;
void insert(Point a) {
	while(top > 1 && ((q[top] - q[top - 1]) ^ (a - q[top - 1])) < 0) top--;
	q[++top] = a;
}
void GetConHull(vector<Point> &v) {
	sort(v.begin(), v.end());
	q[++top] = v[0];
	for(int i = 1; i < v.size(); i++) if(v[i] != v[i - 1]) insert(v[i]);
	for(int i = v.size() - 2; i >= 0; i--) if(v[i] != v[i + 1]) insert(v[i]);
	v.clear(); 
	for(int i = 1; i <= top; i++) v.push_back(q[i]); top = 0;
}
void Merge(vector<Point> &a, vector<Point> &b) {
	vector<Point> c;
	q[++top] = a[0] + b[0]; 
	int i = 0, j = 0;
	while(i + 1 < a.size() && j + 1< b.size()) {
		Point n1 = (a[i] + b[j + 1]) - q[top], n2 = (a[i + 1] + b[j]) - q[top];
		if((n1 ^ n2) < 0) 
			q[++top] = a[i + 1] + b[j], i++;
		else 
			q[++top] = a[i] + b[j + 1], j++; 
	}
	for(; i < a.size(); i++) q[++top] = a[i] + b[b.size() - 1];
	for(; j < b.size(); j++) q[++top] = b[j] + a[a.size() - 1];
	for(int i = 1; i <= top; i++) c.push_back(q[i]);
	LL ans = 0;
	//for(auto &g : c) printf("%d %d
", g.x, g.y);
	for(int i = 1; i < c.size() - 1; i++) 
		ans += (c[i] - c[0]) ^ (c[i + 1] - c[0]);
	cout << ans;
}
signed main() {
	N = read(); M = read();
	for(int i = 1; i <= N; i++) {
		int x = read(), y = read();
		v1.push_back({x, y});
	}
	for(int i = 1; i <= M; i++) {
		int x = read(), y = read();
		v2.push_back({x, y});
	}
	GetConHull(v1);
	GetConHull(v2);
	Merge(v1, v2);
	return 0;
}
/*
4 5
0 0 2 1 0 1 2 0
0 0 1 0 0 2 1 2 0 1
*/