USACO 3.4 分析——2011年3月19日

     这一节TEXT介绍的是计算几何的内容。关于计算几何的内容，还要分开来具体阐述，难点是凸包。

第一题：

题目大意：逆时针给定若干个点，判断这些点能否构成一个多边形，并且给定一个观察点，求此点可观察到的线段

算法：枚举+计算几何

这道题目很纠结，花了半天时间研究MAIGO的代码，大致理解了，但并不能完全消化。

/*ID: zhangji42TASK: fence4LANG: C++NOTE: This program is from maigo.*/#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;const int maxn = 200;const double pi = atan(1)*4;const double zero = 1e-6;//#define max(a,b) (a > b?a : b)//#define min(a,b) (a < b?a : b)int x[maxn], y[maxn];int n, px, py;double a[maxn], b[maxn];bool vis[maxn];int main(){ freopen("fence4.in", "r", stdin); freopen("fence4.out", "w", stdout); int next(int); int scalar(int,int,int,int,int,int); int cross(int,int,int,int,int,int); int sgn(int); int intersect(int,int,int,int,int,int,int,int); bool anycross(); double angle(int,int); double distance(double,double,double, double); void look(double); scanf("%d%d%d", &n, &px, &py); for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]); if (anycross()) printf("NOFENCE/n"); else { /************************************************* **use the polar system to sort the angle **we use vis[] to mark the seeing of a segment **use the Perpendicular bisector of two vector from view point to find the seeing segment **the ouput form should be same as the input order with the second point first, then the first point **************************************************/ for (int i = 1; i <= n; i++) a[i] = angle(x[i]-px, y[i]-py); memcpy(b, a, sizeof(a)); sort(b+1, b+n+1); memset(vis,0,sizeof(vis)); for (int i = 2; i <= n; i++) if (b[i] - b[i-1] > zero) look((b[i]+b[i-1])/2); double t = (b[n]+b[1]-2*pi)/2; if (t < 0) look(2*pi+t); else look(t); int cnt = 0; for (int i = 1; i <= n; i++) cnt += vis[i]; printf("%d/n", cnt); for (int i = 1; i <= n-2; i++) if (vis[i]) printf("%d %d %d %d/n", x[i], y[i], x[i+1], y[i+1]); if (vis[n]) printf("%d %d %d %d/n", x[1], y[1], x[n], y[n]); if (vis[n-1]) printf("%d %d %d %d/n", x[n-1], y[n-1], x[n], y[n]); } return 0;}/*point to the next point especially for x = n with 1 next */int next(int x) { if (x < n) return x+1; else return 1;}int sgn(int x) { if (x > 0) return 1; if (x == 0) return 0; if (x < 0) return -1;}/*get the dot-product at the start point (xa, ya) */int scalar(int xa, int ya, int xb, int yb, int xc, int yc) { int x1 = xb-xa, y1 = yb-ya; int x2 = xc-xa, y2 = yc-ya; return x1 * x2 + y1 * y2;}/*get the cross-product at the start point(xa, ya), the direction from ab -> ac */int cross(int xa, int ya, int xb, int yb, int xc, int yc) { int x1 = xb-xa, y1 = yb-ya; int x2 = xc-xa, y2 = yc-ya; return x1 * y2 - x2 * y1;}/*judge whether the four point intersect each other inside*/bool intersect(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) { return (sgn(cross(x1, y1, x2, y2, x3, y3)) * sgn(cross(x1, y1, x2, y2, x4, y4)) < 0) && (sgn(cross(x3, y3, x4, y4, x1, y1)) * sgn(cross(x3, y3, x4, y4, x2, y2)) < 0);}/*get the Euclidean distance*/double distance(double x1, double y1, double x2, double y2) { return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));}/*judge whether it can form a fence*/bool anycross() { for (int i = 1; i <= n-2; i++) for (int j = i+2; j <= n; j++) if (intersect(x[i], y[i], x[next(i)], y[next(i)], x[j], y[j], x[next(j)], y[next(j)])) return true; return false;}/*get the angle from tangent */double angle(int x, int y) { if (x == 0) { if (y > 0) return pi/2; else return pi*1.5;//tangent not exist } else { double t = atan(1.0*y/x);//NOTICE THE TYPE!!!! if (x > 0) { if (y > 0) return t; else return pi*2+t; } else return t+pi; }}void look(double angle) { double min = 1e30, alpha, beta; int x1, y1, x2, y2, v = 0; for (int i = 1; i <= n; i++) { /********************************************************************** **(x1, y1) is in the lower position, (x2, y2) is in the higher position **alpha is the angle of (x1, y1), beta is the angle of (x2, y2) **make sure that angle between alpha and beta **In the triangle ABC with D in a BC, S(ABD)/S(ACD) = Sin(beta-angle) * AD * AB / (Sin(angle - alpha) * AD * AC) = DB / DC = t **the position of D can be worked out with (x1, y1) and (x2, y2),so the distance of PD is known **we get the minimal distance as the first seeing line ***********************************************************************/ if (cross(px, py, x[i], y[i], x[next(i)], y[next(i)]) > 0) { x1 = x[i]; y1 = y[i]; x2 = x[next(i)]; y2 = y[next(i)]; alpha = a[i]; beta = a[next(i)]; } else { x1 = x[next(i)]; y1 = y[next(i)]; x2 = x[i]; y2 = y[i]; alpha = a[next(i)]; beta = a[i]; } if (sin(angle - alpha) > 0 && sin(beta - angle) > 0) { double t = sin(angle - alpha) / sin(beta - angle) * distance(px, py, x1, y1) / distance(px, py, x2, y2); double d = distance(px, py, (x1+t*x2)/(1+t), (y1+t*y2)/(1+t)); if (d < min) { min = d; v = i; } } } if (min < 1e30) vis[v] = true;}

 

第二题：American Heritage

算法：中序遍历

/*ID: zhangji42TASK: heritageLANG: C++**learn to traversal the tree in inoder**string.find(char/string):int (the start position, 0 as start)**string.substr(start,len):string**string.size():int*/#include <iostream>#include <string>#include <stdio.h>using namespace std;string s1, s2; void print(string s1, string s2) { if (s1.size() >= 1) { int k = s1.find(s2[0]); print(s1.substr(0, k), s2.substr(1, k)); print(s1.substr(k+1, s1.size()-k-1), s2.substr(k+1, s1.size()-k-1)); cout << s2[0]; } }int main(){ freopen("heritage.in", "r", stdin); freopen("heritage.out", "w", stdout); cin >> s1 >> s2; print(s1, s2); cout << endl; return 0;}

 

第三题：Electric Fence

算法：皮克定理

/*ID: zhangji42TASK: fence9LANG: C++皮克定律： S = A + B/2 -1，其中S为三角形的面积，A为内部点数，B为线段上点数*/#include <stdio.h>#include <string.h>int gcd(int n, int m) { if (m == 0) return n; if (n < 0) n = -n; if (m < 0) m = -m; int r = n % m; while (r) { n = m; m = r; r = n % m; } return m;}int getonline(int x, int y) { return gcd(x, y)+1;}int main(){ freopen("fence9.in", "r", stdin); freopen("fence9.out", "w", stdout); int n, m, p, s; scanf("%d %d %d", &n, &m, &p); s = m*p/2+1; s -= (getonline(n,m) + (p+1) + getonline(n-p, m) -3)/2; printf("%d/n", s); return 0;}

 

第四题：Raucaus Rockers

算法1：枚举 + 位运算 + 剪枝

#include <stdio.h>#include <string.h>const int maxn = 21;int l[maxn];int n, t, m;int main(){ freopen("rockers.in", "r", stdin); freopen("rockers.out", "w", stdout); scanf("%d %d %d", &n, &t, &m); for (int i = 0; i < n; i++) scanf("%d", &l[i]); int ii = 1 << n; int max = 0; for (int i = 0; i < ii; i++) { int nowleft = t, mleft = m-1, now = 0; for (int j = 0; j < n; j++) { if (now + n-j < max) break;//最优性剪枝 if (i & (1 << (n-j-1))) { now++; if (l[j] <= nowleft) { nowleft -= l[j]; } else { mleft--; nowleft = t-l[j]; if (nowleft < 0 || mleft < 0) break;//可行性剪枝 } } } if (now > max && nowleft >= 0 && mleft >= 0) max = now; } printf("%d/n", max); return 0;}

算法2：DP（经典）

以下来子NOCOW网站上的写得很好的一个程序，值得学习

/* // ID:nettle99 PROG:rockers LANG:C++ */ #include <cstdio> #define MAXN 21 FILE *fin = fopen("rockers.in", "r"); FILE *fout = fopen("rockers.out", "w"); int max[MAXN], g[MAXN][MAXN], t[MAXN]; int main() { int N, T, M; fscanf(fin, "%d%d%d", &N, &T, &M); for (int i = 1; i <= N; i++) fscanf(fin, "%d", &t[i]); for (int i = 1; i <= N; i++) for (int j = M; j >= 1; j--) { g[j][0] = max[j - 1]; for (int v = T; v >= t[i]; v--) { if (g[j][v] <= g[j][v - t[i]] + 1) g[j][v] = g[j][v - t[i]] + 1; if (g[j][v] > max[j]) max[j] = g[j][v]; } } fprintf(fout, "%d/n", max[M]); return 0; }

 

 

**DP的题目思路不是很清晰，只会基本的形式，有必要认真研读DD_ENGI的《背包九讲》。

**计算几何的功底很薄弱，需要针对性加强