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实现的功能如下——在一个N个点的无环图中,共有N-1条边,M个访问中每次询问两个点的距离
原理——既然N个点,N-1条边,则说明这是一棵树,而且联通。所以以1为根节点DFS建树,然后通过求两点的LCA的方式,先求得最近公共祖先,然后再通过深度来求出两点距离
1 type 2 point=^node; 3 node=record 4 g:longint; 5 next:point; 6 end; 7 const 8 maxn=100500; 9 maxm=trunc(ln(maxn)/ln(2))+1; 10 var 11 i,j,k,l,m,n:longint; 12 a:array[0..maxn] of point; 13 c:array[0..maxm,0..maxn] of longint; 14 d:array[0..maxn] of longint; 15 function max(x,y:longint):longint;inline; 16 begin 17 if x>y then max:=x else max:=y; 18 end; 19 function min(x,y:longint):longint;inline; 20 begin 21 if x<y then min:=x else min:=y; 22 end; 23 procedure swap(var x,y:longint);inline; 24 var z:longint; 25 begin 26 z:=x;x:=y;y:=z; 27 end; 28 procedure add(x,y:longint);inline; 29 var p:point; 30 begin 31 new(p);p^.g:=y; 32 p^.next:=a[x];a[x]:=p; 33 end; 34 procedure dfs(x:longint);inline; 35 var p:point; 36 begin 37 p:=a[x]; 38 while p<>nil do 39 begin 40 if c[0,p^.g]=0 then 41 begin 42 c[0,p^.g]:=x; 43 d[p^.g]:=d[x]+1; 44 dfs(p^.g); 45 end; 46 p:=p^.next; 47 end; 48 end; 49 function getfat(x,y:longint):longint;inline; 50 var i,j,k:longint; 51 begin 52 i:=0; 53 while y>0 do 54 begin 55 if odd(y) then x:=c[i,x]; 56 inc(i);y:=y div 2; 57 end; 58 exit(x); 59 end; 60 function getcom(x,y:longint):longint;inline; 61 var i:longint; 62 begin 63 if d[x]<d[y] then swap(x,y); 64 x:=getfat(x,d[x]-d[y]); 65 if x=y then exit(x); 66 for i:=maxm downto 0 do 67 begin 68 if c[i,x]<>c[i,y] then 69 begin 70 x:=c[i,x]; 71 y:=c[i,y]; 72 end; 73 end; 74 exit(c[0,x]); 75 end; 76 77 begin 78 readln(n,m); 79 for i:=1 to n do a[i]:=nil; 80 for i:=1 to n-1 do 81 begin 82 readln(j,k); 83 add(j,k);add(k,j); 84 end; 85 fillchar(d,sizeof(d),0); 86 fillchar(c,sizeof(c),0); 87 c[0,1]:=-1; 88 dfs(1); 89 for i:=1 to maxm do 90 for j:=1 to n do 91 c[i,j]:=c[i-1,c[i-1,j]]; 92 for i:=1 to m do 93 begin 94 readln(j,k); 95 l:=getcom(j,k); 96 writeln(d[j]+d[k]-d[l]-d[l]); 97 end; 98 readln; 99 end. 100
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原文地址:http://www.cnblogs.com/HansBug/p/4235072.html