学了ZZY的算法,就要过一下他出的题。
题目大意:
给出一些直线,求半平面交的面积。
解题思路:
半平面交求面积。
下面是代码:
#include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> #define eps 1e-10 #define pi acos(-1.0) #define inf 107374182 #define inf64 1152921504606846976 #define lc l,m,tr<<1 #define rc m + 1,r,tr<<1|1 #define zero(a) fabs(a)<eps #define iabs(x) ((x) > 0 ? (x) : -(x)) #define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE)) #define clearall(A, X) memset(A, X, sizeof(A)) #define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE)) #define memcopyall(A, X) memcpy(A , X ,sizeof(X)) #define max( x, y ) ( ((x) > (y)) ? (x) : (y) ) #define min( x, y ) ( ((x) < (y)) ? (x) : (y) ) using namespace std; const int maxn = 30005; int dq[maxn], top, bot, pn, order[maxn], ln,num; struct Point { double x, y; } p[maxn]; struct Line { Point a, b; double angle; } l[maxn]; int dblcmp(double k) { if(fabs(k) < eps) return 0; return k > 0 ? 1 : -1; } double multi(Point p0, Point p1, Point p2) { return (p1.x-p0.x)*(p2.y-p0.y) - (p1.y-p0.y)*(p2.x-p0.x); } bool cmp(int u, int v) { int d = dblcmp(l[u].angle-l[v].angle); if(!d) return dblcmp(multi(l[u].a, l[v].a, l[v].b)) > 0; //大于0取向量左半部分为半平面,小于0,取右半部分 return d < 0; } void getIntersect(Line l1, Line l2, Point& p) { double dot1, dot2; dot1 = multi(l2.a, l1.b, l1.a); dot2 = multi(l1.b, l2.b, l1.a); p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1); p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1); } bool judge(Line l0, Line l1, Line l2) { Point p; getIntersect(l1, l2, p); return dblcmp(multi(p, l0.a, l0.b)) < 0; //大于小于符号与上面cmp()中注释处相反 } void addLine(double x1, double y1, double x2, double y2) { l[ln].a.x = x1; l[ln].a.y = y1; l[ln].b.x = x2; l[ln].b.y = y2; l[ln].angle = atan2(y2-y1, x2-x1); order[ln] = ln; ln++; } void halfPlaneIntersection() { int i, j; sort(order, order+ln, cmp); for(i = 1, j = 0; i < ln; i++) if(dblcmp(l[order[i]].angle-l[order[j]].angle) > 0) order[++j] = order[i]; ln = j + 1; dq[0] = order[0]; dq[1] = order[1]; bot = 0; top = 1; for(i = 2; i < ln; i++) { while(bot < top && judge(l[order[i]], l[dq[top-1]], l[dq[top]])) top--; while(bot < top && judge(l[order[i]], l[dq[bot+1]], l[dq[bot]])) bot++; dq[++top] = order[i]; } while(bot < top && judge(l[dq[bot]], l[dq[top-1]], l[dq[top]])) top--; while(bot < top && judge(l[dq[top]], l[dq[bot+1]], l[dq[bot]])) bot++; num=0; for(i=bot; i<top; i++) { getIntersect(l[dq[i]],l[dq[i+1]],p[num++]); } if(top > bot+1)getIntersect(l[dq[top]], l[dq[bot]],p[num++]); } double xmul(Point p0,Point p1,Point p2) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } double Get_area() { double area = 0; for(int i = 1; i < num-1; i++) area += xmul(p[0], p[i], p[i+1]); return fabs(area)/2.0; } int main() { int i; double x1,x2,y1,y2; scanf ("%d", &pn); for(ln = i = 0; i < pn; i++) { scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); addLine(x1, y1, x2,y2); } addLine(0,0,10000,0); addLine(10000,0,10000,10000); addLine(10000,10000,0,10000); addLine(0,10000,0,0); halfPlaneIntersection(); printf("%.1f\n",Get_area()); return 0; }
POJ 2451 Uyuw's Concert
时间: 2024-11-08 21:49:27