UVa 1331 最大面积最小的三角剖分

左手的ㄟ右手 2021-12-18 14:41 235阅读 0赞

## 题意：输入一个多边形（不一定是凸包），找一个最大三角形面积最小的三角剖分，输出最大三角形的面积。 ##

## 分析：最优三角剖分。**dp\[i\]\[j\]表示从i点到j点的最优值,枚举中间点k**。**转移方程为dp\[i\]\[j\]=min(dp\[i\]\[j\],max(area(i,j,k),max(dp\[i\]\[k\],dp\[k\]\[j\])))**。但是题目中给的不一定是凸多边形，则在对任意一个子多边形，如果有任意一个对角线i, j 和多边形中的任意一条边相交，则设置d(i, j)为 无穷大即可，边界是d(i, i + 1) = 0。 ##

## 参考：[https://www.cnblogs.com/Konjakmoyu/p/4905563.html][https_www.cnblogs.com_Konjakmoyu_p_4905563.html]。 ##

## 代码： ##

#include<bits/stdc++.h>
    using namespace std;
    #define INF 0xfffffff
    const int Maxm=50+5;
    
    int m;
    int x[Maxm],y[Maxm];
    double d[Maxm][Maxm];
    
    double area(int a,int b,int c)
    {
        double s=(double)(1.0/2)*(x[a]*y[b]+x[b]*y[c]+x[c]*y[a]-x[a]*y[c]-x[b]*y[a]-x[c]*y[b]);
        if(s<0) return -s;
        return s;
    }
    ///面积法判断三角形是否合法（a-b是否为对角线）
    bool check(int a,int b,int c)
    {
        int i;
        for(i=1;i<=m;i++)
        {
            if(i==a||i==b||i==c) continue;
            double d=area(a,b,i)+area(a,c,i)+area(b,c,i)-area(a,b,c);
            if(d<0) d=-d;
            if(d<=0.01) return 0;
        }
        return 1;
    }
    
    int main()
    {
        int n;
        scanf("%d",&n);
        while(n--)
        {
            int i,j,k;
            scanf("%d",&m);
            for(i=1;i<=m;i++) scanf("%d%d",&x[i],&y[i]);
            for(i=m;i>=1;i--)
            {
                d[i][i+1]=0.0;
                for(j=i+2;j<=m;j++)
                {
                    d[i][j]=INF;
                    for(k=i+1;k<j;k++)
                    {
                        if(check(i,j,k)) ///判断i j k三点构成的三角形是否合法（三角形内不包含其他点）
                        d[i][j]=min(d[i][j],max(max(area(i,j,k),d[i][k]),d[k][j]));
                    }
                }
            }
            printf("%.1lf\n",d[1][m]);
        }
        return 0;
    }

## 刘汝佳代码： ##

#include<cstdio>
    #include<cstdlib>
    #include<cmath>
    #include<cstring>
    #include<vector>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const double eps = 1e-10;
    int dcmp(double x)
    {
        if(fabs(x) < eps) return 0;
        else return x < 0 ? -1 : 1;
    }
    
    struct Point
    {
        double x, y;
        Point(double x=0, double y=0):x(x),y(y) { }
    };
    
    typedef Point Vector;
    
    Vector operator + (const Vector& A, const Vector& B)
    {
        return Vector(A.x+B.x, A.y+B.y);
    }
    Vector operator - (const Point& A, const Point& B)
    {
        return Vector(A.x-B.x, A.y-B.y);
    }
    Vector operator * (const Vector& A, double p)
    {
        return Vector(A.x*p, A.y*p);
    }
    
    bool operator < (const Point& a, const Point& b)
    {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    
    bool operator == (const Point& a, const Point &b)
    {
        return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
    }
    
    double Dot(const Vector& A, const Vector& B)
    {
        return A.x*B.x + A.y*B.y;
    }
    double Cross(const Vector& A, const Vector& B)
    {
        return A.x*B.y - A.y*B.x;
    }
    double Length(Vector A)
    {
        return sqrt(Dot(A, A));
    }
    
    bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
    {
        double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
               c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
        return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
    }
    
    bool OnSegment(const Point& p, const Point& a1, const Point& a2)
    {
        return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
    }
    
    typedef vector<Point> Polygon;
    
    int isPointInPolygon(const Point& p, const Polygon& poly)
    {
        int n = poly.size();
        int wn = 0;
        for(int i = 0; i < n; i++)
        {
            const Point& p1 = poly[i];
            const Point& p2 = poly[(i+1)%n];
            if(p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1; // 在边界上
            int k = dcmp(Cross(p2-p1, p-p1));
            int d1 = dcmp(p1.y - p.y);
            int d2 = dcmp(p2.y - p.y);
            if(k > 0 && d1 <= 0 && d2 > 0) wn++;
            if(k < 0 && d2 <= 0 && d1 > 0) wn--;
        }
        if (wn != 0) return 1; // 内部
        return 0; // 外部
    }
    
    const int maxn = 100 + 5;
    
    bool isDiagonal(const Polygon& poly, int a, int b)
    {
        int n = poly.size();
        for(int i = 0; i < n; i++)
            if(i != a && i != b && OnSegment(poly[i], poly[a], poly[b])) return false; // 中间不能有其他点
        for(int i = 0; i < n; i++)
            if(SegmentProperIntersection(poly[i], poly[(i+1)%n], poly[a], poly[b])) return false; // 不能和多边形的边规范相交
        Point midp = (poly[a] + poly[b]) * 0.5;
        return (isPointInPolygon(midp, poly) == 1); // 整条线段在多边形内
    }
    
    const double INF = 1e9;
    double d[maxn][maxn];
    
    double solve(const Polygon& poly)
    {
        int n = poly.size();
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++) d[i][j] = -1;
        for(int i = n-2; i >= 0; i--)
            for(int j = i+1; j < n; j++)
            {
                if(i + 1 == j) d[i][j] = 0;
                else if(!(i == 0 && j == n-1) && !isDiagonal(poly, i, j)) d[i][j] = INF;
                else
                {
                    d[i][j] = INF;
                    for(int k = i+1; k < j; k++)
                    {
                        double m = max(d[i][k], d[k][j]);
                        double area = fabs(Cross(poly[j]-poly[i], poly[k]-poly[i])) / 2.0; // triangle i-j-k
                        m = max(m, area);
                        d[i][j] = min(d[i][j], m);
                    }
                }
            }
        return d[0][n-1];
    }
    
    int main()
    {
        int T, n;
        scanf("%d", &T);
        while(T--)
        {
            scanf("%d", &n);
            double x, y;
            Polygon poly;
            for(int i = 0; i < n; i++)
            {
                scanf("%lf%lf", &x, &y);
                poly.push_back(Point(x,y));
            }
            printf("%.1lf\n", solve(poly));
        }
        return 0;
    }

[https_www.cnblogs.com_Konjakmoyu_p_4905563.html]: https://www.cnblogs.com/Konjakmoyu/p/4905563.html