poj 1264 || UVA 109 SCUD Busters

UVA这题过得蛮多人的，觉得应该是个水题吧。

 

一读题，也不麻烦，给你N个王国，求下凸包，再求面积。给你一些炮弹，问炮弹炸掉的面积。（一个炮弹炸的话，整个王国都被炸了）。

 

直接求凸包后，求出各个王国的面积，然后判断炮弹在哪个王国里，这个直接用判断点是否在多边形内。

 

UVA WA了一次，如果多个炮弹打到同一个王国里，面积只算一次的，改了后，A掉了。

 

poj上一搜，居然也有这题。。才100+人过 = =。。。其实套套模板就过去了，不过套得时候很没信心啊。。。那么长。。

 

#include <queue> #include <stack> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h> #include <string.h> #include <string> #include <algorithm> using namespace std; const int MAX = 110; struct point{ double x,y;}; //点 struct polygon{ point c[MAX],a; double area; int n;}; struct segment{ point a,b;}; // 线段 const double eps = 1e-6; bool dy(double x,double y) { return x > y + eps;} // x > y bool xy(double x,double y) { return x < y - eps;} // x < y bool dyd(double x,double y) { return x > y - eps;} // x >= y bool xyd(double x,double y) { return x < y + eps;} // x <= y bool dd(double x,double y) { return fabs( x - y ) < eps;} // x == y polygon king[MAX]; point c[MAX]; double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 { return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y); } double disp2p(point a,point b) { return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) ); } double area_polygon(point p[],int n) { double s = 0.0; for(int i=0; i<n; i++) s += p[(i+1)%n].y * p[i].x - p[(i+1)%n].x * p[i].y; return fabs(s)/2.0; } bool cmp(point a,point b) // 排序 { double len = crossProduct(c[0],a,b); if( dd(len,0.0) ) return xy(disp2p(c[0],a),disp2p(c[0],b)); return xy(len,0.0); } bool onSegment(point a, point b, point c) { double maxx = max(a.x,b.x); double maxy = max(a.y,b.y); double minx = min(a.x,b.x); double miny = min(a.y,b.y); if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,minx) && xyd(c.x,maxx) && dyd(c.y,miny) && xyd(c.y,maxy) ) return true; return false; } bool segIntersect(point p1,point p2, point p3, point p4) { double d1 = crossProduct(p3,p4,p1); double d2 = crossProduct(p3,p4,p2); double d3 = crossProduct(p1,p2,p3); double d4 = crossProduct(p1,p2,p4); if( xy(d1 * d2,0.0) && xy(d3 * d4,0.0) ) return true; if( dd(d1,0.0) && onSegment(p3,p4,p1) ) return true;//如果不判端点相交，则下面这四句话不需要 if( dd(d2,0.0) && onSegment(p3,p4,p2) ) return true; if( dd(d3,0.0) && onSegment(p1,p2,p3) ) return true; if( dd(d4,0.0) && onSegment(p1,p2,p4) ) return true; return false; } bool point_inPolygon(point pot,point p[],int n) { int count = 0; segment l; l.a = pot; l.b.x = 1e10*1.0; l.b.y = pot.y; p[n] = p[0]; for(int i=0; i<n; i++) { if( onSegment(p[i],p[i+1],pot) ) return true; if( !dd(p[i].y,p[i+1].y) ) { int tmp = -1; if( onSegment(l.a,l.b,p[i]) ) tmp = i; else if( onSegment(l.a,l.b,p[i+1]) ) tmp = i+1; if( tmp != -1 && dd(p[tmp].y,max(p[i].y,p[i+1].y)) ) count++; else if( tmp == -1 && segIntersect(p[i],p[i+1],l.a,l.b) ) count++; } } if( count % 2 == 1 ) return true; return false; } point stk[MAX]; int top; double Graham(int n) { int tmp = 0; for(int i=1; i<n; i++) if( xy(c[i].x,c[tmp].x) || dd(c[i].x,c[tmp].x) && xy(c[i].y,c[tmp].y) ) tmp = i; swap(c[0],c[tmp]); sort(c+1,c+n,cmp); stk[0] = c[0]; stk[1] = c[1]; top = 1; for(int i=2; i<n; i++) { while( xyd( crossProduct(stk[top],stk[top-1],c[i]), 0.0 ) && top >= 1 ) top--; stk[++top] = c[i]; } return area_polygon(stk,top+1); } int main() { int n; int i = 0; double x,y; while( ~scanf("%d",&n) && n != -1 ) { king[i].n = n; for(int k=0; k<n; k++) scanf("%lf %lf",&king[i].c[k].x,&king[i].c[k].y); king[i].a = king[i].c[0]; i++; } double sum = 0.0; for(int k=0; k<i; k++) { memcpy(c,king[k].c,sizeof(king[k].c)); king[k].area = Graham(king[k].n); memcpy(king[k].c,stk,sizeof(stk)); king[k].n = top+1; } point pot; bool die[MAX]; memset(die,false,sizeof(die)); while( ~scanf("%lf %lf",&pot.x,&pot.y) ) { for(int k=0; k<i; k++) if( point_inPolygon(pot,king[k].c,king[k].n) && !die[k] ) { die[k] = true; sum += king[k].area; } } printf("%.2lf/n",sum); return 0; }