uva 11123 - Counting Trapizoid(容斥+几何)

题目链接:uva 11123 - Counting Trapizoid

题目大意:给定若干个点,问有多少种梯形,不包括矩形,梯形的面积必须为正数。因为是点的集合,所以不会优重复的点。

解题思路:枚举两两点,求出该条直线,包括斜率k,偏移值c,以及长度l。已知梯形的性质,一对对边平行,也就是说一对平行但是不相等的边。

所以将所有线段按照斜率排序,假设对于某一斜率,有m条边,那么这m条边可以组成的含平行对边的四边形有C(2m),要求梯形还要减掉长度相同以及共线的情况,分别对应的是l相同和c相同,但是根据容斥原理,要加回l和c均相等的部分。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;
const int N = 205;
const double eps = 1e-9;
const double pi = 4 * atan(1.0);

struct line {
    double k, c, l;
    line (double k, double c = 0, double l = 0) {
        this->k = k;
        this->c = c;
        this->l = l;
    }
    friend bool operator < (const line& a, const line& b) {
        return a.k < b.k;
    }
};

struct state {
    double c, l;
    state (double c, double l) {
        this->c = c;
        this->l = l;
    }
};

int n;
double x[N], y[N];
vector<line> set;
vector<state> g;

inline double distant (double xi, double yi) {
    return xi * xi + yi * yi;
}

inline double cal (int a, int b) {
    if (x[a] == x[b])
        return x[a];
    return y[a] - x[a] * ( (y[a] - y[b]) / (x[a] - x[b]) );
}

inline int C (int a) {
    if (a < 1)
        return 0;
    return a * (a - 1) / 2;
}

inline bool cmpC (const state& a, const state& b) {
    if (fabs(a.c-b.c) > eps)
        return a.c < b.c;
    return a.l < b.l;
}

inline bool cmpL (const state& a, const state& b) {
    return a.l < b.l;
}

int judge () {
    sort(g.begin(), g.end(), cmpC);
    int ans = 0, cnt = 1, tmp = 1;

    /*
    for (int i = 0; i < g.size(); i++) {
        printf("%lf %lf\n", g[i].c, g[i].l);
    }
    printf("\n");
    */

    for (int i = 1; i < g.size(); i++) {
        if (fabs(g[i].c - g[i-1].c) > eps) {
            ans = ans + C(cnt) - C(tmp);
            cnt = 0;
            tmp = 0;
        }

        if (fabs(g[i].l - g[i-1].l) > eps) {
            ans = ans - C(tmp);
            tmp = 0;
        }

        tmp++;
        cnt++;
    }
    ans = ans + C(cnt) - C(tmp);

    sort(g.begin(), g.end(), cmpL);
    cnt = 1;
    for (int i = 1; i < g.size(); i++) {
        if (fabs(g[i].l - g[i-1].l) > eps) {
            ans = ans + C(cnt);
            cnt = 0;
        }
        cnt++;
    }
    ans = ans + C(cnt);
    return ans;
}

int solve () {

    int ans = 0;

    if (set.size() == 0)
        return 0;

    g.clear();
    g.push_back(state(set[0].c, set[0].l));

    for (int i = 1; i < set.size(); i++) {

        //printf("%d %lf!!!!!!!\n", i, set[i].k);
        if (fabs(set[i].k - set[i-1].k) > eps) {
            ans += C(g.size()) - judge();
            g.clear();
        }
        g.push_back(state(set[i].c, set[i].l));
    }

    ans += C(g.size()) - judge();
    return ans;
}

int main () {
    int cas = 1;
    while (scanf("%d", &n) == 1 && n) {

        set.clear();
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &x[i], &y[i]);
            for (int j = 0; j < i; j++)
                set.push_back(line( atan2(y[j]-y[i], x[j]-x[i]), cal(i, j), distant(y[i]-y[j], x[i]-x[j]) ));
        }

        for (int i = 0; i < set.size(); i++) {
            if (set[i].k < eps)
                set[i].k += pi;
        }

        sort(set.begin(), set.end());

        /*
        for (int i = 0; i < set.size(); i++) {
            printf("%d %lf %lf %lf\n", i, set[i].k, set[i].c, set[i].l);
        }
        */

        printf("Case %d: %d\n", cas++, solve());
    }
    return 0;
}

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uva 11123 - Counting Trapizoid(容斥+几何)

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