100道动态规划——4 UVA 1331 Maximum Triangulation DP 计算几何

                  好吧，这道题在练练DP的同时顺便抄了一下刘汝佳dalao的板子，其实我感觉重点是计算几何？23333

                  按照紫书上的顺序，这道题就是线性DP的例题的终结了，接下来是树型DP了。

                  定义Dp[i][j]表示从第i个顶点到第j个顶点所能产生的最小的最大三角形，然后从i~j里面枚举k。

                  把dalao的代码重构了一下

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

using namespace std;
const double eps=1e-10;

int dcmp(double x){return (fabs(x)<eps)?0:(x<0?-1:1);}

struct Point{
double x,y;
Point(double a=0,double b=0):x(a),y(b){}
};

Point operator +(const Point& a,const Point& b){return Point(a.x+b.x,a.y+b.y);}
Point operator -(const Point& a,const Point& b){return Point(a.x-b.x,a.y-b.y);}
Point operator *(const Point& a,double p){return Point(a.x*p,a.y*p);}
Point operator *(double x,const Point& a){return a*x;}

bool operator <(const Point& a,const Point& b){return a.x<b.x||dcmp(a.x-b.x)==0&&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 Point& a,const Point& b){return a.x*b.x+a.y*b.y;}
double Cross(const Point& a,const Point& b){return a.x*b.y-a.y*b.x;}
double Length(const Point& a){return sqrt(Dot(a,a));}

bool SegmentProperIntersection(const Point& a1,const Point& a2,const Point& b1,const Point& b2); //判断线段a1a2和线段b1b2是不是相交
bool OnSegment(const Point& p,const Point& a1,const Point& a2); //判断点p是不是在a1和a2的连线上

typedef vector<Point> Polygon;

int isPointInPolygon(const Point& p,const Polygon& poly); //判断p是不是在poly内
bool isDiagonal(const Polygon& poly,int a,int b); //判断顶点a和顶点b构成的连线算不算对角线
double solve(const Polygon& poly);

int T,m;
double dp[110][110];

int main(){
scanf("%d",&T);
Polygon poly;
while(T--){
scanf("%d",&m);
double x,y;
for(int i=0;i<m;++i)
scanf("%lf%lf",&x,&y),poly.push_back(Point(x,y));

printf("%.1lf\n",solve(poly));
poly.clear();
}
return 0;
}

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;
}

int isPointInPolygon(const Point& p,const Polygon& poly){
int n=poly.size(),wn=0;
for(int i=0;i<n;++i){
const Point& p1=poly[i],& p2=poly[(i+1)%n];
if(p1==p||p2==p||OnSegment(p,p1,p2))
return -1;
int k=dcmp(Cross(p2-p1,p-p1)),d1=dcmp(p1.y-p.y),d2=dcmp(p2.y-p.y);
if(k>0&&d1<=0&&d2>0)++wn;
if(k<0&&d2<=0&&d1>0)--wn;
}
return wn?1:0;
}

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;
for(int i=0;i<n;++i)
if(SegmentProperIntersection(poly[i],poly[(i+1)%n],poly[a],poly[b]))
return false;

return (isPointInPolygon((poly[a]+poly[b])*0.5,poly)==1);
}

double solve(const Polygon& poly){
int n=poly.size();
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
dp[i][j]=-1;

for(int i=n-2;i>=0;--i)
for(int j=i+1;j<n;++j){
if(i+1==j)
dp[i][j]=0;
else if(!(i==0&&j==n-1)&&!isDiagonal(poly,i,j))
dp[i][j]=DBL_MAX;
else{
dp[i][j]=DBL_MAX;
for(int k=i+1;k<j;++k)
dp[i][j]=min(dp[i][j],max(max(dp[i][k],dp[k][j]),fabs(Cross(poly[j]-poly[i],poly[k]-poly[i]))/2.0));
}
}

return dp[0][n-1];
}