由于k只有2，所以我们分类讨论
显然当k=1时，我们只要连一条最长的路径即可就是树的直径L
少走了L-1条边
如果k=2时，我们再次连边成环后
如果成环路径与上一次的最长路径没有相同的边，那少走的边数是路径长l-1
如果有相同的边，那么相同的边一共还是会走两次，少走的边数是l-1-2*same
因此我们只要把第一次找的的最长路径上的边标记为-1，再做一次树形dp即可
1 type node=record 2 po,len,next:longint; 3 end; 4 5 var w:array[0..200010] of node; 6 f,p1,p2,p:array[0..100010] of longint; 7 v:array[0..100010] of boolean; 8 i,ans,loc,maxx,t,n,k,x,y:longint; 9 10 function max(a,b:longint):longint; 11 begin 12 if a>b then exit(a) else exit(b); 13 end; 14 15 procedure add(x,y:longint); 16 begin 17 inc(t); 18 w[t].po:=y; 19 w[t].len:=1; 20 w[t].next:=p[x]; 21 p[x]:=t; 22 end; 23 24 procedure dfs(x:longint); 25 var i,y,s1,s2:longint; 26 begin 27 i:=p[x]; 28 v[x]:=true; 29 s1:=0; 30 s2:=0; 31 while i<>-1 do 32 begin 33 y:=w[i].po; 34 if not v[y] then 35 begin 36 dfs(y); 37 if f[y]+w[i].len>s1 then 38 begin 39 s2:=s1; 40 p2[x]:=p1[x]; 41 s1:=f[y]+w[i].len; 42 p1[x]:=i; 43 end 44 else if (s2<f[y]+w[i].len) then 45 begin 46 s2:=f[y]+w[i].len; 47 p2[x]:=i; 48 end; 49 end; 50 i:=w[i].next; 51 end; 52 // writeln(x,' ',s1,' ',s2); 53 f[x]:=s1; 54 if maxx<s1+s2 then 55 begin 56 maxx:=s1+s2; 57 loc:=x; 58 end; 59 end; 60 61 begin 62 t:=-1; 63 fillchar(p,sizeof(p),255); 64 fillchar(p1,sizeof(p1),255); 65 fillchar(p2,sizeof(p2),255); 66 readln(n,k); 67 for i:=1 to n-1 do 68 begin 69 readln(x,y); 70 add(x,y); 71 add(y,x); 72 end; 73 dfs(1); 74 ans:=2*(n-1)-maxx+1; 75 if k=2 then 76 begin 77 maxx:=0; 78 w[p1[loc]].len:=-1; 79 w[p2[loc]].len:=-1; 80 x:=w[p1[loc]].po; 81 while p1[x]<>-1 do 82 begin 83 w[p1[x]].len:=-1; 84 x:=w[p1[x]].po; 85 end; 86 x:=w[p2[loc]].po; 87 while p1[x]<>-1 do 88 begin 89 w[p1[x]].len:=-1; 90 x:=w[p1[x]].po; 91 end; 92 fillchar(v,sizeof(v),false); 93 fillchar(f,sizeof(f),0); 94 dfs(1); 95 ans:=ans-maxx+1; 96 end; 97 writeln(ans); 98 end.
