给你两条线及两条线上的点，求最小生成树。
可以挨个枚举一条线上的点，三分出另一条线上离他最近的点进行连边。
注意N、M可能为0
debug了1天半，至今不知道原始二分版本错在哪里。。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<vector>
7 #include<cmath>
8 #include<queue>
9 #include<set>
10 using namespace std;
11 #define N 50010
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 struct point
18 {
19 double x,y;
20 int id;
21 point(double x=0,double y = 0):x(x),y(y) {}
22 } p[N],q[N];
23 struct node
24 {
25 int u,v;
26 double w;
27 } ed[N<<4];
28 int fa[N],g;
29 double t1[N],t2[N];
30 point a,b,c,d;
31 typedef point pointt;
32 point operator -(point a,point b)
33 {
34 return point(a.x-b.x,a.y-b.y);
35 }
36 int dcmp(double x)
37 {
38 if(fabs(x)<eps) return 0;
39 return x<0?-1:1;
40 }
41
42 double dis(point a)
43 {
44 return sqrt(a.x*a.x+a.y*a.y);
45 }
46 bool cmmp(node ta,node tb)
47 {
48 return ta.w<tb.w;
49 }
50 int findx(int x)
51 {
52 if(x!=fa[x])
53 fa[x] = findx(fa[x]);
54 return fa[x];
55 }
56 void add(point ta,point tb)
57 {
58 ed[++g].u = ta.id;
59 ed[g].v = tb.id;
60 ed[g].w = dis(ta-tb);
61 }
62 int main()
63 {
64 // freopen("data.in","r",stdin);
65 // freopen("data.out","w",stdout);
66 int t,i,n,m;
67 int kk = 0;
68 cin>>t;
69 while(t--)
70 {
71 scanf("%d%d",&n,&m);
72 scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);
73 for(i = 0; i <= n+m ; i++)
74 fa[i] = i;
75 for(i = 0; i < n; i++)
76 {
77 scanf("%lf",&t1[i]);
78
79 // cout<<p[i].x<<" "<<p[i].y<<endl;
80 }
81 for(i = 0; i< m; i++)
82 {
83 scanf("%lf",&t2[i]);
84
85 }
86 sort(t1,t1+n);
87 sort(t2,t2+m);
88 int h=0;
89 for(i=0;i<n;i++)
90 if(i==n-1||t1[i]!=t1[i+1])
91 t1[h++]=t1[i];
92 n=h;
93 h=0;
94 for(i=0;i<m;i++)
95 if(i==m-1||t2[i]!=t2[i+1])
96 t2[h++]=t2[i];
97 m=h;
98 for(i = 0; i < n;i++)
99 {
100 p[i] = point(a.x*t1[i]+b.x*(1-t1[i]),a.y*t1[i]+b.y*(1-t1[i]));
101 p[i].id = i+1;
102 }
103 for(i = 0 ; i < m ; i++)
104 {
105 q[i] = point(c.x*t2[i]+d.x*(1-t2[i]),c.y*t2[i]+d.y*(1-t2[i]));
106 q[i].id = i+n+1;
107 }
108 g = 0;
109 double sum1 = 0,sum2 = 0;
110 for(i = 0 ; i < n-1 ; i++)
111 {
112 add(p[i],p[i+1]);
113 sum1+=dis(p[i]-p[i+1]);
114 }
115 for(i = 0 ; i < m-1 ; i++)
116 {
117 add(q[i],q[i+1]);
118 sum2+=dis(q[i]-q[i+1]);
119 }
120 if(n==0||m==0)
121 {
122
123 printf("Case #%d: %.3lf\n",++kk,sum1+sum2);
124 continue;
125 }
126 for(i = 0; i < n; i++)
127 {
128 // point pp ;
129 // if(dcmp(c.x-d.x)==0)
130 // pp = point(c.x,p[i].y);
131 // else
132 // pp = dispoint(c,d,p[i]);
133 // if(dcmp(pp.x-min(c.x,d.x))<=0||dcmp(pp.x-max(c.x,d.x))>=0)
134 // {
135 // add(i,n+1);
136 // if(n+2<n+m)
137 // add(i,n+2);
138 // if(n+m-1>n+1)
139 // add(i,n+m-1);
140 // add(i,n+m);
141 // continue;
142 // }
143 // int k = find(n+1,n+m,c,dis(pp-c));
144 // cout<<p[k].id<<" "<<pp.x<<" "<<pp.y<<endl;
145 // add(i,k);
146 // if(k!=n+1)
147 // add(i,k-1);
148 // if(k!=n+m)
149 // add(i,k+1);
150 int l, r;
151 l = 0;
152 r = m-1;
153 while (r - l > 1)
154 {
155 int mid1 = (r + l) >> 1;
156 int mid2 = (mid1 + r) >> 1;
157 if (dis(p[i]-q[mid1]) > dis(p[i]-q[mid2]))
158 l = mid1;
159 else
160 r = mid2;
161 }
162 // int k = find(1,n,a,dis(pp-a));
163 add(p[i],q[l]);
164 add(p[i],q[r]);
165 if(l-1>=0)
166 add(p[i],q[l-1]);
167 if(r+1<=m-1)
168 add(p[i],q[r+1]);
169 }
170 sort(ed+1,ed+g+1,cmmp);
171 int num = 0;
172 double sum = 0;
173 for(i = 1; i <= g; i++)
174 {
175 int tx = findx(ed[i].u);
176 int ty = findx(ed[i].v);
177 if(tx!=ty)
178 {
179 fa[tx] = ty;
180 num++;
181 sum+=ed[i].w;
182 }
183 }
184 printf("Case #%d: %.3f\n",++kk,sum);
185 }
186 return 0;
187 }
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时间： 2024-08-26 16:14:00