定义: 对于欧式空间上的两个点集 $A, B$, 存在点集 $C$ 满足 ${a + b \in C|a \in A,b\in B }$ ($a, b$均为向量的形式), 则称 $C$ 为$A, B$的闵可夫斯基和, 即 $$ C = A + B $$

差 (different): $C = A - B = {a - b | a \in A, b \in B} = A + (-B)$

表示对$\forall a \in A, \forall b \in B, \exists c \in C (c = a - b)$

或者可以这样理解$\forall c \in C, \exists b \in B, \exists a \in A (a = c + b)$

==warning==: 这里的加号和减号不是一对可以互逆的运算, $(A - B) + B \neq A$

wikipedia

  • 例题(第一道过的洛谷黑题):

战争

题目大意: 有两个部落$A,B$, 两个部落分别有$n, m$个人, 如两个部落所围成的凸包没有重合的点, 则不会发生战争. 现在$B$部落的全部人打算向$(x, y)$向量迁移, 有$q$次询问, 问是否会发生战争.

思路是找到一个点集$C$, 满足对 $\forall c \in C, \exists b \in B, c + b \in A$, 这和前面差的定义是相同的, 故此题转化为了求$C = A - B$, 然后每次询问二分判断点是否在$C$内.

参考代码 (包含了闵可夫斯基和以及判断一个点是否在凸包内):

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#include<bits/stdc++.h>
using namespace std;
using Real = long long;
struct Point {
    Real x, y;
    Point() {}
    Point(Real x, Real y) : x(x), y(y) {}
    friend istream& operator >> (istream &is, Point& v) {
        return is >> v.x >> v.y;
    }
    Point& operator+= (const Point& p) {
        x += p.x, y += p.y;
        return *this;
    }
    Point& operator-= (const Point& p) {
        x -= p.x, y -= p.y;
        return *this;
    }
    Point operator+ (const Point& p) const {
        return Point(*this) += p;
    }
    Point operator- (const Point& p) const {
        return Point(*this) -= p;
    }
    Point operator- () {
        return Point(-x, -y);
    }
    friend Real crs(const Point& a, const Point& b) {
        return a.x * b.y - a.y * b.x;
    }
    bool operator< (const Point& p) const {
        if(x != p.x) return x < p.x;
        return y < p.y;
    }
};
using Points = vector<Point>;
enum position { CCW = 1, CW = -1, ON = 0 };
int ccw(const Point& a, Point b, Point c) {
    b -= a, c -= a;
    if(crs(b, c) > 0) return CCW;
    if(crs(b, c) < 0) return CW;
    return ON;
}
int ccw(Point a, Point b) {
    return ccw(Point(0, 0), a, b);
}
Points Convexhell(Points& ps) {
    int n = ps.size(), k = 0;
    Points res(2 * n);
    sort(ps.begin(), ps.end());
    for(int i = 0; i < n; res[k ++] = ps[i ++]) {
        while(k >= 2 && ccw(res[k - 2], res[k - 1], ps[i]) != CCW) {
            k --;
        }
    }
    for(int i = n - 2, t = k + 1; i >= 0; res[k ++] = ps[i --]) {
        while(k >= t && ccw(res[k - 2], res[k - 1], ps[i]) != CCW) {
            k --;
        }
    }
    res.resize(k - 1);
    return res;
}
Points Minkowski(Points& A, Points& B) {
    int n = A.size(), m = B.size();
    Points res (n + m + 1);
    res[0] = A[0] + B[0];
    int k = 1, i = 0, j = 0;
    int cA = 0, cB = 0;
    while(cA < n && cB < m) {
        int nxi = (i + 1) % n, nxj = (j + 1) % m;
        Point x = A[nxi] - A[i];
        Point y = B[nxj] - B[j];
        if(ccw(x, y) == CCW) {
            res[k ++] = res[k - 1] + x;
            i = nxi, cA ++;
        } else if(ccw(x, y) == CW) {
            res[k ++] = res[k - 1] + y;
            j = nxj, cB ++;
        } else {
            res[k ++] = res[k - 1] + x + y;
            j = nxj, i = nxi;
            cA ++, cB ++;
        }
    }
    while(cA < n) {
        int nxi = (i + 1) % n;
        res[k ++] = res[k - 1] + A[nxi] - A[i];
        cA ++, i = nxi;
    }
    while(cB < m) {
        int nxj = (j + 1) % m;
        res[k ++] = res[k - 1] + B[nxj] - B[j];
        cB ++, j = nxj;
    }
    res.resize(k - 1);
    return res;
}
bool check_in(Point& p, Points& ps) {
    int n = ps.size();
    int O = 0, l = 1, r = n - 1;
    if(ccw(ps[O], ps[r], p) == CCW || ccw(ps[O], ps[l], p) == CW) {
        return false;
    }
    while(l + 1 < r) {
        int m = l + r >> 1;
        if(ccw(ps[O], ps[m], p) == CCW) {
            l = m;
        } else {
            r = m;
        }
    }
    if(ccw(ps[l], ps[r], p) == CW) {
        return false;
    } else {
        return true;
    }
}
void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    Points A(n), B(m);
    for(int i = 0; i < n; i ++) {
        cin >> A[i];
    }
    for(int i = 0; i < m; i ++) {
        cin >> B[i];
        B[i] = -B[i];
    }
    A = Convexhell(A); n = A.size();
    B = Convexhell(B); m = B.size();
    Points C = Minkowski(A, B);
    while(q --) {
        Point query;
        cin >> query;
        if(check_in(query, C)) {
            cout << "1\n";
        } else {
            cout << "0\n";
        }
    }
}
int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int _ = 1;
    while(_ --) solve();
}