例题9-11 最大面积的最小的三角剖分 UVa1331

1.题目描述：点击打开链接

2.解题思路：本题属于“最优三角剖分”型的dp，设d(i,j)表示子多边形i,i+1,...,j-1(i<j)的最优解，则不难列出状态转移方程如下：

d(i,j)=min(S(i,j,k),max(d(i,k),d(k,j))|i<k<j);

其中，S(i,j,k)为三角形i-j-k的面积。不过此时需要保证i-j是对角线（唯一的例外是i=0且j=n-1），具体做法是当边i-j不满足条件时直接设为INF，其他部分和凸多边形的情形完全一样。

3.代码：

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<cassert>
#include<string>
#include<sstream>
#include<set>
#include<bitset>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<functional>
using namespace std;

#define me(s)  memset(s,0,sizeof(s))
#define rep(i,n) for(int i=0;i<(n);i++)
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair <int, int> P;

const double eps=1e-10;

int dcmp(double x)
{
    if(fabs(x)<eps)return 0;
    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 Vector&a,const Vector&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);
    double 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)return 1; //点在内部
    return 0;  //点在外部
}

bool isDiagnal(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; //线段中间不能有其他点
    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;
const int N=100+10;
double d[N][N];

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)&&!isDiagnal(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; //三角形i-j-k的面积
                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));
    }
}