景驰无人驾驶 1024 编程邀请赛 B题 计算几何+裸二分匹配

暴力n2建边，然后跑二分图匹配，比赛时候写了一个BUG代码，调了比赛一个半小时，赛后半小时才过。

在check的时候，我是直接求出每个矩形四个顶点，然后矩形面积交求答案。

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
const double pi = acos(-1.0);
#define inf 0x3f3f3f3f
using namespace std;
struct scana{
    int x,y,v,a,xita,w,l;
}A[200];
struct scanb{
    int x,y,xita,w,l;
}B[200];
/*
 * 多边形的交，多边形的边一定是要按逆时针方向给出
 * 还要判断是凸包还是凹包，调用相应的函数
 * 面积并，只要和面积减去交即可
 */
const int maxn = 20;
const double eps = 1e-6;
int dcmp(double x)
{
    if(x > eps) return 1;
    return x < -eps ? -1 : 0;
}
struct Point
{
    double x, y;
};
double cross(Point a,Point b,Point c) ///叉积
{
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
Point intersection(Point a,Point b,Point c,Point d)
{
    Point p = a;
    double t =((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
    p.x +=(b.x-a.x)*t;
    p.y +=(b.y-a.y)*t;
    return p;
}
//计算多边形面积
double PolygonArea(Point p[], int n)
{
    if(n < 3) return 0.0;
    double s = p[0].y * (p[n - 1].x - p[1].x);
    p[n] = p[0];
    for(int i = 1; i < n; ++ i)
        s += p[i].y * (p[i - 1].x - p[i + 1].x);
    return fabs(s * 0.5);
}
double CPIA(Point a[], Point b[], int na, int nb)//ConvexPolygonIntersectArea
{
    Point p[20], tmp[20];
    int tn, sflag, eflag;
    a[na] = a[0], b[nb] = b[0];
    memcpy(p,b,sizeof(Point)*(nb + 1));
    for(int i = 0; i < na && nb > 2; i++)
    {
        sflag = dcmp(cross(a[i + 1], p[0],a[i]));
        for(int j = tn = 0; j < nb; j++, sflag = eflag)
        {
            if(sflag>=0) tmp[tn++] = p[j];
            eflag = dcmp(cross(a[i + 1], p[j + 1],a[i]));
            if((sflag ^ eflag) == -2)
                tmp[tn++] = intersection(a[i], a[i + 1], p[j], p[j + 1]); ///求交点
        }
        memcpy(p, tmp, sizeof(Point) * tn);
        nb = tn, p[nb] = p[0];
    }
    if(nb < 3) return 0.0;
    return PolygonArea(p, nb);
}
double SPIA(Point a[], Point b[], int na, int nb)///SimplePolygonIntersectArea 调用此函数
{
    int i, j;
    Point t1[4], t2[4];
    double res = 0, num1, num2;
    a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0];
    for(i = 2; i < na; i++)
    {
        t1[1] = a[i-1], t1[2] = a[i];
        num1 = dcmp(cross(t1[1], t1[2],t1[0]));
        if(num1 < 0) swap(t1[1], t1[2]);
        for(j = 2; j < nb; j++)
        {
            t2[1] = b[j - 1], t2[2] = b[j];
            num2 = dcmp(cross(t2[1], t2[2],t2[0]));
            if(num2 < 0) swap(t2[1], t2[2]);
            res += CPIA(t1, t2, 3, 3) * num1 * num2;
        }
    }
    return res;
}
Point p1[maxn], p2[maxn];
double a,b,c;
double dis(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2) + (y1 -y2)*(y1-y2));
}
bool check(int i,int j){

    double s1 = 1.0*A[i].l*A[i].w;
    double s2 = 1.0*B[j].l*B[j].w;
    double det = a*1.0*A[i].v + a*1.0*a*A[i].a/2;
    double xita = A[i].xita*1.0/180*pi;
    double x1 = A[i].x*1.0 + det * cos(xita);
    double y1 = A[i].y*1.0 + det * sin(xita);
    double x2 = B[j].x*1.0;
    double y2 = B[j].y*1.0;

    double ss = sqrt(1.0*A[i].w * 1.0*A[i].w /4.0 + 1.0*A[i].l * 1.0*A[i].l /4.0);
    double sxita = atan(A[i].w*1.0/A[i].l);

    p1[0].x = x1 + ss * cos(xita - sxita);
    p1[0].y = y1 + ss * sin(xita - sxita);

    p1[1].x = x1 + ss * cos(xita + sxita);
    p1[1].y = y1 + ss * sin(xita + sxita);

    p1[2].x = x1 - ss * cos(xita - sxita);
    p1[2].y = y1 - ss * sin(xita - sxita);

    p1[3].x = x1 - ss * cos(xita + sxita);
    p1[3].y = y1 - ss * sin(xita + sxita);

    xita = B[j].xita*1.0/180*pi;
    ss = sqrt(1.0*B[j].w * 1.0*B[j].w /4.0 + 1.0*B[j].l * 1.0*B[j].l /4.0);
    sxita = atan(B[j].w*1.0/B[j].l);

    p2[0].x = x2 + ss * cos(xita - sxita);
    p2[0].y = y2 + ss * sin(xita - sxita);

    p2[1].x = x2 + ss * cos(xita + sxita);
    p2[1].y = y2 + ss * sin(xita + sxita);

    p2[2].x = x2 - ss * cos(xita - sxita);
    p2[2].y = y2 - ss * sin(xita - sxita);

    p2[3].x = x2 - ss * cos(xita + sxita);
    p2[3].y = y2 - ss * sin(xita + sxita);
    double Area = SPIA(p1, p2, 4, 4);
    if (Area>=(s1+s2-Area) * b || dis(x1,y1,x2,y2) <= c){
        return true;
    }
    return false;
}
const int MAXN=300;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的
int linker[MAXN];
bool used[MAXN];
bool dfs(int u)
{
    int v;
    for(v=0;v<vN;v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<uN;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}
int main()
{
    int  T;
    scanf("%d",&T);
    while (T--){
        int n,m;
        scanf("%d%d",&n,&m);
        cin >> a >> b >> c;
        for (int i = 1; i<=n; i++){
            scanf("%d%d%d%d%d%d%d",&A[i].x,&A[i].y,&A[i].v,&A[i].a,&A[i].xita,&A[i].l,&A[i].w);
        }
        for (int i = 1; i<=m; i++){
            scanf("%d%d%d%d%d",&B[i].x,&B[i].y,&B[i].xita,&B[i].l,&B[i].w);
        }
        memset(g,0,sizeof(g));
        uN = n;
        vN = m;
        for (int i = 1; i<=n;i++){
            for (int j = 1; j<=m; j++){
                if (check(i,j)){
                    g[i-1][j-1] = 1;
                }
            }
        }
        int ans = hungary();
        cout << ans << endl;
    }
    return 0;
}