树分治入门
poj1741是男人八题之一,经典的树分治的题目
这里用到的是点分治
核心思想是我们把某个点i作为根,把路径分为过点i和不过点i
先统计过点i这样的路径数,然后在统计其子树中的答案,这样就不断地划分成规模较小子问题。
要使划分最优,我们每次都选的是树的重心
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 type node=record 2 len,next,po:longint; 3 end; 4 5 var e:array[0..80010] of node; 6 w,p,d,a,size:array[0..40010] of longint; 7 v:array[0..40010] of boolean; 8 root,sum,n,t,ans,k,i,x,y,z,len: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 sort(l,r: longint); 16 var i,j,x,y: longint; 17 begin 18 i:=l; 19 j:=r; 20 x:=a[(l+r) div 2]; 21 repeat 22 while a[i]<x do inc(i); 23 while x<a[j] do dec(j); 24 if not(i>j) then 25 begin 26 y:=a[i]; 27 a[i]:=a[j]; 28 a[j]:=y; 29 inc(i); 30 j:=j-1; 31 end; 32 until i>j; 33 if l<j then sort(l,j); 34 if i<r then sort(i,r); 35 end; 36 37 procedure add(x,y,z:longint); 38 begin 39 inc(len); 40 e[len].po:=y; 41 e[len].len:=z; 42 e[len].next:=p[x]; 43 p[x]:=len; 44 end; 45 46 procedure getroot(x,fa:longint); //选择重心 47 var i,y:longint; 48 begin 49 i:=p[x]; 50 size[x]:=1; 51 w[x]:=0; 52 while i<>0 do 53 begin 54 y:=e[i].po; 55 if (not v[y]) and (y<>fa) then 56 begin 57 getroot(y,x); 58 size[x]:=size[x]+size[y]; 59 w[x]:=max(w[x],size[y]); 60 end; 61 i:=e[i].next; 62 end; 63 w[x]:=max(w[x],sum-size[x]); 64 if w[x]<w[root] then root:=x; 65 end; 66 67 procedure deep(x,fa:longint); 68 var i,y:longint; 69 begin 70 inc(t); 71 a[t]:=d[x]; 72 i:=p[x]; 73 while i<>0 do 74 begin 75 y:=e[i].po; 76 if not v[y] and (y<>fa) then 77 begin 78 d[y]:=d[x]+e[i].len; 79 deep(y,x); 80 end; 81 i:=e[i].next; 82 end; 83 end; 84 85 function calc(x,now:longint):longint; 86 var i,l,r:longint; 87 begin 88 t:=0; 89 d[x]:=now; 90 calc:=0; 91 deep(x,0); 92 sort(1,t); 93 l:=1; 94 r:=t; 95 while l<r do 96 begin 97 if a[l]+a[r]<=k then 98 begin 99 calc:=calc+r-l; 100 inc(l); 101 end 102 else dec(r); 103 end; 104 end; 105 106 procedure work(x:longint); 107 var i,y:longint; 108 begin 109 ans:=ans+calc(x,0); 110 i:=p[x]; 111 v[x]:=true; 112 while i<>0 do 113 begin 114 y:=e[i].po; 115 if not v[y] then 116 begin 117 ans:=ans-calc(y,e[i].len); //之前的计算会导致lca不为root的点对被算入,实际它们的路径不过重心,这里要减去 118 sum:=size[y]; 119 root:=0; 120 getroot(y,0); 121 work(root); 122 end; 123 i:=e[i].next; 124 end; 125 end; 126 127 begin 128 readln(n,k); 129 while n<>0 do 130 begin 131 len:=0; 132 fillchar(p,sizeof(p),0); 133 for i:=1 to n-1 do 134 begin 135 readln(x,y,z); 136 add(x,y,z); 137 add(y,x,z); 138 end; 139 fillchar(v,sizeof(v),false); 140 sum:=n; 141 w[0]:=2147483647; 142 getroot(1,0); 143 ans:=0; 144 work(root); 145 writeln(ans); 146 readln(n,k); 147 end; 148 end. 149 150
bzoj2152是更水的树分治……