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 1150 绳子围点

题解：

凸包 + 凸包面积 + 皮克定理

 #include <cstdio>
 #include <algorithm>
 using namespace std;
 typedef long long ll;

 const int MAXN =  + ;

 char cj;
 int n, m;

 struct Point{//{{{
     ll x, y;

     friend bool operator < (const Point& A, const Point& B){
         return A.x < B.x || (A.x == B.x && A.y < B.y);
     }
 } p[MAXN], ch[MAXN * ];//}}}

 inline ll xmult(Point a, Point b, Point c){//{{{
     return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
 }

 inline void AndrweScan(){
     sort(p, p + n);
     m = , ch[] = p[], ch[] = p[];
     for (int i = ; i < n; i ++){
         while (m >=  && xmult(ch[m], ch[m - ], p[i]) > )
             m --;
         ch[++ m] = p[i];
     }

     for (int i = n - ; i >= ; i --){
         while (m >=  && xmult(ch[m], ch[m - ], p[i]) > )
             m --;
         ch[++ m] = p[i];
     }
 }//}}}

 inline ll nextLL(){
     bool flag = false;
     do {
         cj = getchar();
         if (cj == '-')
             flag = true;
     } while (!( <= cj && cj <= ));

     ll ret = ;
     do {
         ret = ret *  + cj - ;
         cj = getchar();
     } while ( <= cj && cj <= );

     return flag ? -ret : ret;
 }

 inline int nextInt(){
     do
         cj = getchar();
     while (!( <= cj && cj <= ));

     int ret = ;
     do {
         ret = ret *  + cj - ;
         cj = getchar();
     } while ( <= cj && cj <= );

     return ret;
 }

 inline ll Plot(){
     ll ret = ;
     for (int i = ; i < m; i ++)
         ret += xmult(ch[], ch[i - ], ch[i]);
     return ret;
 }

 inline ll gcd(ll a, ll b){
     return b ? gcd(b, a % b) : a;
 }

 inline ll Cal(int i){
     int j = (i == m - ) ?  : i + ,
         dx = ch[i].x - ch[j].x, dy = ch[i].y - ch[j].y;
     return gcd(abs(dx), abs(dy));
 }

 inline void Pick(){
     ll dS = Plot(), b = , a;
     for (int i = ; i < m; i++)
         b += Cal(i);
     a = (dS - b + ) / ;
     printf("%lld\n", a + b);
 }

 int main(){
     n = nextInt();
     for (int i = ; i < n; i++)
         p[i].x = nextLL(), p[i].y = nextLL();
     AndrweScan();
     Pick();
 }

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

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define INF 100000000
 #define eps 1e-9

 const int MAXN = ;

 int n;
 double dp[MAXN][MAXN];

 struct Point{
     double x, y;

     friend Point operator + (const Point A, const Point B){
         return (Point){A.x+B.x, A.y+B.y};
     }

     friend Point operator - (const Point A, const Point B){
         return (Point){A.x-B.x, A.y-B.y};
     }

     friend Point operator * (const Point A, const Point B){
         return (Point){A.x*B.x, A.y*B.y};
     }

     friend Point operator / (const Point A, const Point B){
         return (Point){A.x/B.x, A.y/B.y};
     }
 }p[MAXN][MAXN];

 inline Point make_Point(double X, double Y){
     return (Point){X, Y};
 }

 inline double len(Point A){
     return sqrt(A.x*A.x+A.y*A.y);
 }

 inline bool In_region(double eg, double l, double r){
     if (l>r) swap(l, r);
     return (l-eps<=eg && eg<=r+eps);
 }

 inline bool Connected(int I1, int I2, int I3, int I4){
     if (I1>I3) swap(I1, I3);
     if ((I1==I3 && I2==I4) || I3-I1==) return true;
     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;
     for (int i=I1+; i<I3; i++){
         bn = p[i][].x*k+b;
         if (!In_region(bn, p[i][].y, p[i][].y) && !In_region(bn, p[i][].y, p[i][].y)) return false;
     }
     return true;
 }

 int main(){
     scanf("%d", &n);

     p[][] = p[][] = p[][] = p[][] = make_Point(, );
     p[n+][] = p[n+][] = p[n+][] = p[n+][] = make_Point(, );

     double x, in, apl;
     for (int i=; i<=n; i++){
         scanf("%lf", &x);
         for (int j=; j<; j++)
             scanf("%lf", &in), p[i][j] = make_Point(x, in);
     }

     //init
     for (int i=; i<; i++)
         dp[][i] = 0.0;
     for (int i=; i<=n+; i++)
         for (int j=; j<; j++)
             dp[i][j] = Connected(, , i, j) ? len(p[i][j]-p[][]) : INF;

     for (int i=; i<=n+; i++)
         for (int j=; j<; j++)
             for (int k=; k<i; k++)
                 for (int l=; l<; l++)
                     if (Connected(k, l, i, j)) dp[i][j] = min(dp[i][j], dp[k][l]+len(p[i][j]-p[k][l]));
     apl = INF;
     for (int i=; i<; i++)
         apl = min(apl, dp[n+][i]);
     printf("%.2lf\n",apl);
     return ;
 }

TYVJ 1462 凸多边形

 #include <cstdio>
 #include <iomanip>
 #include <iostream>
 #include <algorithm>
 #define eps 1e-12
 using namespace std;

 const int MAXN = +;

 int n;

 struct Point{
     long double x, y;

     friend Point operator - (const Point& A, const Point& B){
         return (Point){A.x-B.x, A.y-B.y};
     }

     friend long double operator * (const Point& A, const Point& B){//cross
         return A.x*B.y-A.y*B.x;
     }

     friend bool operator < (const Point& A, const Point& B){
         return (A.x!=B.x) ? A.x<B.x : A.y<B.y;
     }

     inline void print(){
         cout<<setiosflags(ios::fixed)<<setprecision()<<x<<" "<<setiosflags(ios::fixed)<<setprecision()<<y<<"\n";
     }
 }p[MAXN], ch[MAXN];

 inline int dcmp(long double eg){
     if (eg>eps) return ; //>0
     if (eg<-eps) return -;//<0
     else return ;
 }

 inline void GrahamScan(){
     int tot, t;
     sort(p, p+n);

     ch[] = p[], ch[] = p[], tot = ;
     for (int i=; i<n; i++){
         while (tot> && dcmp((ch[tot]-ch[tot-])*(p[i]-ch[tot-]))!=-) tot--;
         ch[++tot] = p[i];
     }

     t = tot, ch[++tot] = p[n-];
     for (int i=n-; i>=; i--){
         while (tot>t && dcmp((ch[tot]-ch[tot-])*(p[i]-ch[tot-]))!=-) tot--;
         ch[++tot] = p[i];
     }

     for (int i=; i<tot; i++)
         ch[i].print();
 }

 int main(){
         scanf("%d", &n);
         for (int i=; i<n; i++)
         cin>>p[i].x>>p[i].y;

     GrahamScan();
     return ;
 }

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

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #define eps 1e-9
 using namespace std;

 const int MAXL = ;

 struct Point{
     double x,y;

     inline void in(){
         char s[MAXL];
         scanf("%s",s);
         int i;
         x = ,y = ;
         for (i=;;i++){
             if (s[i]==',') break;
             x *= ,x += s[i]-;
         }
         for (i+=;;i++){
             if (s[i]==')') break;
             y *= ,y += s[i]-;
         }

     }

     inline void out(){
         printf("%.2lf %.2lf\n",x,y);
     }

     inline bool is_zero(){
         return (y== && x==);
     }

     inline double len(){
         return sqrt(x*x-y*y);
     }

     friend Point operator + (const Point A,const Point B){
         return (Point){A.x+B.x,A.y+B.y};
     }

     friend Point operator - (const Point A,const Point B){
         return (Point){A.x-B.x,A.y-B.y};
     }

     friend Point operator * (const Point A,const double B){
         return (Point){A.x*B,A.y*B};
     }

     friend Point operator / (const Point A,const double B){
         return (Point){A.x/B,A.y/B};
     }

     friend bool operator == (const Point A,const Point B){
         return (A.x==B.x && A.y==B.y);
     }
 }A,B,C,D;

 inline double area(Point p1,Point p2,Point p3){
     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));
 }

 inline bool on(Point Aa,Point Ba,Point q){
     double MX = max(Aa.x,Ba.x);
     double mx = min(Aa.x,Ba.x);
     double MY = max(Aa.y,Ba.y);
     double my = min(Aa.y,Ba.y);
     return ((mx<q.x && q.x<MX) && (my<q.y && q.y<MY)) ? true : false;
 }

 inline bool zero(double eg){
     return (eg>-eps && eg<eps);
 } 

 inline int appleeeeeeee(){
     A.in();B.in();C.in();D.in();
     if (A==D || B==D || C==D) return ;//顶点上
     double s1,s2,s3,s;
     s = area(A,B,C);
     s1 = area(A,B,D);
     s2 = area(A,C,D);
     s3 = area(B,C,D);

     if (zero(s1) && on(A,B,D)) return ;
     if (zero(s2) && on(A,C,D)) return ;
     if (zero(s3) && on(B,C,D)) return ;//边上

     if ((s1+s2+s3-s)<eps && (s1+s2+s3-s)>-eps) return ;
     else return ;
 }

 int main(){
     printf("%d",appleeeeeeee());
     return ;
 }

TYVJ 1544 角平分线

题解：

推个公式就好了

 #include <cstdio>
 #include <cmath>

 struct Point{
     double x,y;

     inline void in(){
         scanf("%lf%lf",&x,&y);
     }

     inline void out(){
         printf("%.2lf %.2lf",x,y);
     }

     friend Point operator + (const Point A,const Point B){
         return (Point){A.x+B.x,A.y+B.y};
     }

     friend Point operator - (const Point A,const Point B){
         return (Point){A.x-B.x,A.y-B.y};
     }

     friend Point operator * (const Point A,const double B){
         return (Point){A.x*B,A.y*B};
     }

     friend Point operator / (const Point A,const double B){
         return (Point){A.x/B,A.y/B};
     }
 }a,b,c,d;

 inline double len(Point eg){
     return sqrt(eg.x*eg.x+eg.y*eg.y);
 }

 int main(){
     a.in();b.in();c.in();
     b = b-a,c = c-a;
     double AB = len(b),AC = len(c);
     double cj = (AB*AC)/(AB+AC);
     d = b/AB + c/AC;
     d = d*cj;
     d = d + a;
     d.out();
     return ;
 }