CodeForces 166B（凸包）

来源：互联网发布：java工程师面试题100 编辑：程序博客网时间：2024/04/30 23:57
判断一下A所构成的凸包是不是跟总的凸包一样就可以了..
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <set>using namespace std;const int INF = ~0U >> 1;const int maxn = 200000 + 7;const int mod = 1000000000 + 7;const double eps = 1e-6;typedef long long LL;struct point {    double x, y;//    bool operator<(const point &rhs) const {//        if(x != rhs.x) return x < rhs.x;//        return y < rhs.y;//    }    bool operator==(const point &rhs) const {        if(fabs(x - rhs.x) < eps || fabs(y - rhs.y) < eps) return true;        return false;    }    bool operator!=(const point &rhs) const {        if(fabs(x - rhs.x) > eps || fabs(y - rhs.y) > eps) return true;        return false;    }};bool mult(point sp, point ep, point op) {    return (sp.x - op.x) * (ep.y - op.y) > (ep.x - op.x) * (sp.y - op.y);}bool operator < (const point &l, const point &r) {    return l.y < r.y || (l.y == r.y && l.x < r.x);}int graham(point pnt[], int n, point res[]) {    int i, len, k = 0, top = 1;    sort(pnt, pnt + n);    if (n == 0) return 0;    res[0] = pnt[0];    if (n == 1) return 1;    res[1] = pnt[1];    if (n == 2) return 2;    res[2] = pnt[2];    for (i = 2; i < n; i++) {        while (top && mult(pnt[i], res[top], res[top-1])) top--;        res[++top] = pnt[i];    }    len = top;    res[++top] = pnt[n - 2];    for (i = n - 3; i >= 0; i--) {        while (top!=len && mult(pnt[i], res[top],res[top-1])) top--;        res[++top] = pnt[i];    }    return top;}int n, m;point a[maxn];point ans1[maxn], ans2[maxn];set<point> s;int main() {    scanf("%d", &n);    for(int i = 0; i < n; ++i) {        scanf("%lf%lf", &a[i].x, &a[i].y);        s.insert(a[i]);    }    int res1 = graham(a, n, ans1);    scanf("%d", &m);    bool ok = false;    for(int i = 0; i < m; ++i) {        scanf("%lf%lf", &a[i+n].x, &a[i+n].y);        if(s.count(a[i+n])) {            ok = true;//            cout << i << "====" << endl;        }    }    if(ok) {        printf("NO\n");        return 0;    }//    cout << "==" << endl;    int res2 = graham(a, n + m, ans2);    if(res1 != res2) {        printf("NO\n");        return 0;    }    for(int i = 0; i < res1; ++i) {        if(ans1[i] != ans2[i]) {            printf("NO\n");            return 0;        }    }    printf("YES\n");    return 0;}
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