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     //init
72     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){//cross
20         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; //>0
34     if (eg<-eps) return -1;//<0
35     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 }

TYVJ计算几何