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TYVJ计算几何
今天讲了计算几何,发几道水水的tyvj上的题解...
计算几何好难啊!@Mrs.General....怎么办....
这几道题都是在省选之前做的,所以前面的Point运算啊,dcmp啊,什么什么的,基本上没用,每次都把上次的main()函数删了接着继续写....
原谅我曾经丑出翔的代码...
虽然现在也很丑...
TYVJ 1543 房间最短路
题解:
Floyd,dp[i][j]表示到第i面墙的第j个点的最短路,注意在更新最小值的时候判断两个点的连通
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 #define INF 100000000 7 #define eps 1e-9 8 9 const int MAXN = 1000;10 11 int n;12 double dp[MAXN][MAXN];13 14 struct Point{15 double x, y;16 17 friend Point operator + (const Point A, const Point B){18 return (Point){A.x+B.x, A.y+B.y};19 }20 21 friend Point operator - (const Point A, const Point B){22 return (Point){A.x-B.x, A.y-B.y};23 }24 25 friend Point operator * (const Point A, const Point B){26 return (Point){A.x*B.x, A.y*B.y};27 }28 29 friend Point operator / (const Point A, const Point B){30 return (Point){A.x/B.x, A.y/B.y};31 }32 }p[MAXN][MAXN];33 34 inline Point make_Point(double X, double Y){35 return (Point){X, Y};36 }37 38 inline double len(Point A){39 return sqrt(A.x*A.x+A.y*A.y);40 }41 42 inline bool In_region(double eg, double l, double r){43 if (l>r) swap(l, r);44 return (l-eps<=eg && eg<=r+eps);45 }46 47 inline bool Connected(int I1, int I2, int I3, int I4){48 if (I1>I3) swap(I1, I3);49 if ((I1==I3 && I2==I4) || I3-I1==1) return true;50 double k = (p[I1][I2].y-p[I3][I4].y)/(p[I1][I2].x-p[I3][I4].x), b = p[I1][I2].y-p[I1][I2].x*k, bn;51 for (int i=I1+1; i<I3; i++){52 bn = p[i][0].x*k+b;53 if (!In_region(bn, p[i][0].y, p[i][1].y) && !In_region(bn, p[i][2].y, p[i][3].y)) return false; 54 }55 return true;56 }57 58 int main(){59 scanf("%d", &n);60 61 p[0][0] = p[0][1] = p[0][2] = p[0][3] = make_Point(0, 5);62 p[n+1][0] = p[n+1][1] = p[n+1][2] = p[n+1][3] = make_Point(10, 5);63 64 double x, in, apl;65 for (int i=1; i<=n; i++){66 scanf("%lf", &x);67 for (int j=0; j<4; j++)68 scanf("%lf", &in), p[i][j] = make_Point(x, in); 69 }70 71 //init72 for (int i=0; i<4; i++)73 dp[0][i] = 0.0;74 for (int i=1; i<=n+1; i++)75 for (int j=0; j<4; j++)76 dp[i][j] = Connected(0, 0, i, j) ? len(p[i][j]-p[0][0]) : INF;77 78 for (int i=0; i<=n+1; i++)79 for (int j=0; j<4; j++)80 for (int k=0; k<i; k++)81 for (int l=0; l<4; l++)82 if (Connected(k, l, i, j)) dp[i][j] = min(dp[i][j], dp[k][l]+len(p[i][j]-p[k][l]));83 apl = INF;84 for (int i=0; i<4; i++) 85 apl = min(apl, dp[n+1][i]);86 printf("%.2lf\n",apl);87 return 0;88 }
TYVJ 1462 凸多边形
1 #include <cstdio> 2 #include <iomanip> 3 #include <iostream> 4 #include <algorithm> 5 #define eps 1e-12 6 using namespace std; 7 8 const int MAXN = 100001+10; 9 10 int n;11 12 struct Point{13 long double x, y;14 15 friend Point operator - (const Point& A, const Point& B){16 return (Point){A.x-B.x, A.y-B.y};17 }18 19 friend long double operator * (const Point& A, const Point& B){//cross20 return A.x*B.y-A.y*B.x;21 }22 23 friend bool operator < (const Point& A, const Point& B){24 return (A.x!=B.x) ? A.x<B.x : A.y<B.y;25 }26 27 inline void print(){28 cout<<setiosflags(ios::fixed)<<setprecision(4)<<x<<" "<<setiosflags(ios::fixed)<<setprecision(4)<<y<<"\n";29 }30 }p[MAXN], ch[MAXN];31 32 inline int dcmp(long double eg){33 if (eg>eps) return 1; //>034 if (eg<-eps) return -1;//<035 else return 0;36 }37 38 inline void GrahamScan(){39 int tot, t;40 sort(p, p+n);41 42 ch[0] = p[0], ch[1] = p[1], tot = 1;43 for (int i=2; i<n; i++){44 while (tot>0 && dcmp((ch[tot]-ch[tot-1])*(p[i]-ch[tot-1]))!=-1) tot--;45 ch[++tot] = p[i];46 }47 48 t = tot, ch[++tot] = p[n-2];49 for (int i=n-3; i>=0; i--){50 while (tot>t && dcmp((ch[tot]-ch[tot-1])*(p[i]-ch[tot-1]))!=-1) tot--;51 ch[++tot] = p[i];52 }53 54 for (int i=0; i<tot; i++)55 ch[i].print();56 }57 58 int main(){59 scanf("%d", &n); 60 for (int i=0; i<n; i++)61 cin>>p[i].x>>p[i].y;62 63 GrahamScan();64 return 0;65 }
TYVJ 1523 神秘大三角
题解:
这道题考的是读入...面积随便整一下就好了
1 #include <cstdio> 2 #include <cmath> 3 #include <algorithm> 4 #define eps 1e-9 5 using namespace std; 6 7 const int MAXL = 1000; 8 9 struct Point{10 double x,y;11 12 inline void in(){13 char s[MAXL];14 scanf("%s",s);15 int i;16 x = 0,y = 0;17 for (i=1;;i++){18 if (s[i]==‘,‘) break;19 x *= 10,x += s[i]-48;20 }21 for (i+=1;;i++){22 if (s[i]==‘)‘) break;23 y *= 10,y += s[i]-48;24 }25 26 }27 28 inline void out(){29 printf("%.2lf %.2lf\n",x,y);30 }31 32 inline bool is_zero(){33 return (y==0 && x==0);34 }35 36 inline double len(){37 return sqrt(x*x-y*y);38 }39 40 friend Point operator + (const Point A,const Point B){41 return (Point){A.x+B.x,A.y+B.y};42 }43 44 friend Point operator - (const Point A,const Point B){45 return (Point){A.x-B.x,A.y-B.y};46 }47 48 friend Point operator * (const Point A,const double B){49 return (Point){A.x*B,A.y*B};50 }51 52 friend Point operator / (const Point A,const double B){53 return (Point){A.x/B,A.y/B};54 }55 56 friend bool operator == (const Point A,const Point B){57 return (A.x==B.x && A.y==B.y);58 }59 }A,B,C,D;60 61 inline double area(Point p1,Point p2,Point p3){62 return abs(0.5*(p1.x*p2.y+p2.x*p3.y+p3.x*p1.y-p1.y*p2.x-p2.y*p3.x-p3.y*p1.x));63 }64 65 inline bool on(Point Aa,Point Ba,Point q){66 double MX = max(Aa.x,Ba.x);67 double mx = min(Aa.x,Ba.x);68 double MY = max(Aa.y,Ba.y);69 double my = min(Aa.y,Ba.y);70 return ((mx<q.x && q.x<MX) && (my<q.y && q.y<MY)) ? true : false;71 }72 73 inline bool zero(double eg){74 return (eg>-eps && eg<eps);75 } 76 77 inline int appleeeeeeee(){78 A.in();B.in();C.in();D.in();79 if (A==D || B==D || C==D) return 4;//顶点上80 double s1,s2,s3,s;81 s = area(A,B,C);82 s1 = area(A,B,D);83 s2 = area(A,C,D);84 s3 = area(B,C,D);85 86 if (zero(s1) && on(A,B,D)) return 3;87 if (zero(s2) && on(A,C,D)) return 3;88 if (zero(s3) && on(B,C,D)) return 3;//边上89 90 if ((s1+s2+s3-s)<eps && (s1+s2+s3-s)>-eps) return 1;91 else return 2;92 }93 94 int main(){95 printf("%d",appleeeeeeee());96 return 0;97 }
TYVJ 1544 角平分线
题解:
推个公式就好了
1 #include <cstdio> 2 #include <cmath> 3 4 struct Point{ 5 double x,y; 6 7 inline void in(){ 8 scanf("%lf%lf",&x,&y); 9 }10 11 inline void out(){12 printf("%.2lf %.2lf",x,y);13 }14 15 friend Point operator + (const Point A,const Point B){16 return (Point){A.x+B.x,A.y+B.y};17 }18 19 friend Point operator - (const Point A,const Point B){20 return (Point){A.x-B.x,A.y-B.y};21 }22 23 friend Point operator * (const Point A,const double B){24 return (Point){A.x*B,A.y*B};25 }26 27 friend Point operator / (const Point A,const double B){28 return (Point){A.x/B,A.y/B};29 }30 }a,b,c,d;31 32 inline double len(Point eg){33 return sqrt(eg.x*eg.x+eg.y*eg.y);34 }35 36 int main(){37 a.in();b.in();c.in();38 b = b-a,c = c-a;39 double AB = len(b),AC = len(c);40 double cj = (AB*AC)/(AB+AC);41 d = b/AB + c/AC;42 d = d*cj;43 d = d + a;44 d.out();45 return 0;46 }
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