[BZOJ1060] [ZJOI2007]时态同步

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=1060

题目大意

给定一棵树及其边权，在边上权值不断+1，使根到所有叶子节点的距离都相同

询问最小+1的次数

题解

树形DP

f[i]:表示以i为根的子树中，根到最深节点的距离f[i]:表示以i为根的子树中，根到最深节点的距离

ans=∑f[i]−f[son[i]−w[i,son[i]]ans=\sum f[i]-f[son[i]-w[i,son[i]]

{$M 100000000,0,100000000}
uses math;
const
maxn=500005;
var
w:array[0..3*maxn,1..3]of longint;
f:array[0..maxn]of longint;
i,j,k:longint;
a,b,c,n,m,len,root:longint;
ans:int64;
procedure init(a,b,c:longint);
begin
w[len,1]:=b; w[len,2]:=c;
if w[a,3]=0 then w[a,3]:=len else w[w[a,1],3]:=len;
w[a,1]:=len; inc(len);
end;

procedure dfs1(a,fa:longint);
var tt:longint;
begin
tt:=w[a,3]; f[a]:=0;
while tt<>0 do
begin
if w[tt,1]<>fa
then begin dfs1(w[tt,1],a); f[a]:=max(f[a],f[w[tt,1]]+w[tt,2]); end;
tt:=w[tt,3];
end;
end;

procedure dfs2(a,fa:longint);
var tt:longint;
begin
tt:=w[a,3];
while tt<>0 do
begin
if w[tt,1]<>fa
then begin ans:=ans+f[a]-f[w[tt,1]]-w[tt,2]; dfs2(w[tt,1],a); end;
tt:=w[tt,3];
end;
end;

begin

readln(n); len:=n+1;
readln(root);
for i:=1 to n-1 do
begin
readln(a,b,c);
init(a,b,c); init(b,a,c);
end;
dfs1(root,0); ans:=0;
dfs2(root,0);
writeln(ans);

end.