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 ",++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 ",++kk,sum); 185 } 186 return 0; 187 }