fzu 2148 Moon Game

题意：判断凸四边形

凸四边形：没有角大于180的四边形。

思路：

不能直接用凸包，要用的话要逆排序点然后使用。

面积法：如果4个点中存在某个点D，Sabd + Sacd + Sbcd = Sabc,则说明是凹四边形。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const double eps = 1e-8;  //定义成double类型

struct point
{
    int x, y;
} p[50];
double area(point a, point b, point c)
{
    return (fabs((b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x))/2);   //一定要fabs(),有可能为负的，
}
bool ok(point a, point b, point c, point d)
{
    if(fabs(area(b,c,d)-area(a,b,c)-area(a,c,d)-area(a,b,d)) < eps)
        return false;
    return true;
}
int main()
{
    int t, i, j, n, k, ans, a, b;
    scanf("%d", &t);
    for(k = 1; k <= t; k++)
    {
        ans = 0;
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%d%d", &p[i].x, &p[i].y);

        if(n<4)
           printf("Case %d: %d", k, ans);
        else
        {
            for(i = 0; i < n; i++)
                for(j = i+1; j < n; j++)
                    for(a = j+1; a < n; a++)
                        for(b = a+1; b < n; b++)
                            if(ok(p[i],p[j],p[a],p[b])&&ok(p[j],p[i],p[a],p[b])
                               &&ok(p[a],p[i],p[j],p[b])&&ok(p[b],p[i],p[j],p[a]))
                                {
                                    ans++;
                                }
            printf("Case %d: %d\n", k, ans);
        }
    }
    return 0;
}

相交法：只要有一对对角线相交，就是凸四边形。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const double eps = 1e-8;  //定义成double类型
class Point
{
public:
    int x, y;
    Point():x(0), y(0) {}
    Point(int xx, int yy): x(xx), y(yy) {}
};
Point p[35];
int n;
 
void init()
{
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> p[i].x >> p[i].y;
}
 
bool judge1(Point a, Point b, Point c, Point d) //判断两条线段是否相交
{
   long long m1 = (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
   long long m2 = (b.x - a.x) * (d.y - a.y) - (d.x - a.x) * (b.y - a.y);
   long long m3 = (d.x - c.x) * (a.y - c.y) - (a.x - c.x) * (d.y - c.y);
   long long m4= (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
    if(m1*m2 <= 0 && m3 *m4 <= 0) return true;
    else return false;
}
bool judge2(int i, int j, int k) //判断3个点是否在一条直线上
{
    int x1 = p[i].x, x2 = p[j].x, x3 = p[k].x;
    int y1 = p[i].y, y2 = p[j].y, y3 = p[k].y;
    if((y2-y1)*(x3-x2) == (y3-y2)*(x2-x1)) return false;
    else return true;
}
int solve()
{
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
        for(int j = i+1; j < n; j++)
        {
            for(int k = j+1; k < n; k++)
            {
                for(int m = k + 1; m < n; m++)
                {
                    if(judge1(p[i], p[j], p[k], p[m])||judge1(p[i], p[m], p[k], p[j]) || judge1(p[i], p[k], p[j], p[m]))
                    cnt++;
                }
            }
        }
    }
    return cnt;
}
int main()
{
    int T;
    scanf("%d", &T);
    for(int i = 1; i <= T; i++)
    {
        init();
        int ans = solve();
        cout << "Case " << i << ": " << ans << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/Jawen/p/10815523.html