Uva 12660 - Ears Cutting

计算几何+dp。要求的是将一个多边形分解成三角形最小的切割长度。先预处理出两点是否可以连线（连线在多边形内），然后再dp。dp[ l ][ r ]表示第l个点到第r个点都切割成三角形需要的最小值。于是就可以写出转移方程。

dp[ l ][ r ] = min(d[ i ][ l ] + d[ i ][ r ] + dp[ l ][ i ] + dp[ i ][ r ] )。枚举第三点i，和l， r 构成三角形。。。然后更新dp[ l ][ r ]就行了。。。因为0， n - 1这两个点是相连的。所以最后答案就是dp[0][n - 1].

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;
const double eps = 1e-6;
const double INF = 1e18;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) { }void read(){scanf("%lf%lf", &x, &y);}
};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (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(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double angle(Vector v) { return atan2(v.y, v.x); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }//点在p线段上
bool OnSegment(Point p, Point a1, Point a2) {return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; //线段包含短点时改成<=
}//线段相交判定
bool SegmentItersection(Point a1, Point a2, Point b1, 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;
}const int N = 110;Point p[N];
int n;
bool c[N][N], vis[N][N];
double dp[N][N], d[N][N];int isPointInPolygon(Point a)
{int wn = 0;for (int i = 0; i < n; i++){if (OnSegment(a, p[i], p[(i + 1) % n])) return -1;    //边界int k = dcmp(Cross(p[(i + 1) % n] - p[i], a - p[i]));int d1 = dcmp(p[i].y - a.y);int d2 = dcmp(p[(i + 1) % n].y - a.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 ok(int a, int b)
{if(!isPointInPolygon((p[a] + p[b]) * 0.5)) return false;for(int i = 0; i < n; i ++){if(SegmentItersection(p[i], p[(i + 1) % n], p[a], p[b])) return false;if(i != a && i != b && OnSegment(p[i], p[a], p[b])) return false;}return true;
}void init()
{for(int i = 0; i < n; i ++){for(int j = 0; j < n; j ++){if(ok(i, j))c[i][j] = true;else c[i][j] = false;d[i][j] = Length(p[i] - p[j]);}}
}double dfs(int l, int r)
{if(vis[l][r]) return dp[l][r];if(r - l <= 2) return 0;vis[l][r] = true;double &ret = dp[l][r];ret = INF;for(int i = l + 2; i <= r - 2; i ++){if(c[l][i] && c[i][r])ret = min(ret, d[i][l] + d[i][r] + dfs(l, i) + dfs(i, r));}if(c[l][r - 1]) ret = min(ret, d[l][r - 1] + dfs(l, r - 1));if(c[l + 1][r]) ret = min(ret, d[l + 1][r] + dfs(l + 1, r));return ret;
}int main()
{int cas = 1;while(scanf("%d", &n) != EOF){for(int i = 0; i < n; i ++){p[i].read();}init();CLR(vis, 0);printf("Case %d: %.4lf\n", cas ++, dfs(0, n - 1));}
}

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