csu1332 割耳法
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割耳法可以把一个多边形切成三角形：每次沿着某条对角线切下来一个三角形（称为“耳朵”），n-3次就能把一个n边形切成一个三角形。如下图，三角形{2,3,4}被割掉了。
输入一个多边形，怎样做才能让每次切割痕迹的总长度最小？
输入最多包含30组测试数据。每组数据第一行为多边形的顶点数n(4<=n<=100)。以下n行描述多边形的各个顶点坐标（均为绝对值不超过10000的整数），按照逆时针或者顺时针排列。
对于每组数据，输出切割痕迹总长度的最小值，保留4位小数。
40 03 01 10 340 010 010 10 1
Case 1: 1.4142Case 2: 10.0499
#include<iostream>#include<stdio.h>#include<cstring>#include<algorithm>#include<cmath>#include<functional>#define eps 1e-9#include<vector>const double INF=1e9;using namespace std;double d[200][200];const double PI = acos(-1.0);int dcmp( double x ){ if( abs(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){}}node[112];typedef point Vector;struct segment{ point a,b; segment(){} segment(point _a,point _b){a=_a,b=_b;}};struct circle{ point c; double r; circle(){} circle(point _c, double _r):c(_c),r(_r) {} point PPP(double a)const{return point(c.x+cos(a)*r,c.y+sin(a)*r);}};struct line{ point p,v; double ang; line() {} line( const point &_p, const point &_v):p(_p),v(_v){ang = atan2(v.y, v.x);} inline bool operator < (const line &L)const{return ang < L.ang;}};point operator + (point a,point b){return point( a.x + b.x,a.y + b.y );}point operator - (point a,point b){return point( a.x - b.x,a.y - b.y );}point operator * (point a,double b){return point( a.x*b,a.y*b );}point operator / (point a,double b){ return point( a.x/b,a.y/b );}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 );}bool operator != (const point &a,const point &b ){return a == b?false:true;}double Dot( point a,point b ){return a.x*b.x + a.y*b.y;} // 点到点的距离 //点积 ab=|a||b|cos@double Length( point a ){return sqrt( Dot( a,a ) );} // 向量长度double Angle( point a,point b ){ return acos( Dot(a,b)/Length(a)/Length(b) );} // 两个向量的角度double D_T_D(const double deg){ return deg/180.0*PI; }// 叉积计算 a*b=|a||b|sin@;double Cross( point a,point b ){ return a.x*b.y - a.y*b.x;}// 线段相交判断 先必须去掉不相交的状态；再判断方向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;}//点pot 是否 在线段 ab 上 只需 叉积等于0 点积等于0bool online(const point a,const point b,const point pot ){ if( Cross( a - pot,b - pot ) == 0 && Dot( a - pot,b - pot ) <= 0 )return 1; return 0;}//p为点,poly为多边形 点在多边形内判定int isinpoly(const point& p, const vector<point> 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 || online(p1, p2 ,p)) return -1; 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}bool isDgnal(const vector<point>poly, int a,int b){ int i,n=poly.size(); for (i=0;i<n;i++) if (i!=a && i!=b && online(poly[a],poly[b],poly[i])) return false; for (i=0;i<n;i++) if (SegmentProperIntersection(poly[i],poly[(i+1)%n],poly[a],poly[b])) return false; point mid=(poly[a]+poly[b])*0.5; return (isinpoly(mid,poly)==1);}double go(const vector<point> poly){ int i,j,k,n=poly.size(); for (i=0;i<n;i++) for (j=0;j<n;j++) d[i][j]=-1; for (i=n-2;i>=0;i--) for (j=i+1;j<n;j++) { double len=Length(poly[i]-poly[j]); if (i+1==j) d[i][j]=0; else if (!(i==0 && j==n-1) && !isDgnal(poly,i,j)) d[i][j]=INF; else { double m=INF; for (k=i+1;k<j;k++) m=min(m,d[i][k]+d[k][j]); d[i][j]=len+m; } } double len0=Length(poly[0]-poly[n-1]); return d[0][n-1]-len0;}int main( ){ int n,cas=1; while (~scanf("%d",&n) && n) { double x,y; vector <point> poly; for (int i=0;i<n;i++) { scanf("%lf%lf",&x,&y); poly.push_back(point(x,y)); } printf("Case %d: %.4lf\n",cas++,go(poly)); }}
