sicily 2377 Ants Colony 【LCA Tarjan算法】
周赛第二次的题目。传说中的最近公共祖先问题(LCA),用Tarjan算法可以过，但注意结果超int.
从题意可以看出这些colony之间构成了一棵树的关系，用Tarjan算法找出每对查询的顶点u,v的公共祖先ancestor,则u,v之间的距离就是dist[u]+dist[v]-2*dist[ancestor],关系显而易见，会者不难。
View Code
#include
<
iostream
>
#include
<
stdio.h
>
#include
<
vector
>
using
namespace
std;
const
int
MAX
=
100001
;
struct
Edge
{
int
end;
int
dis;
Edge(
int
e,
int
d): end(e),dis(d) {}
};
struct
Quer
{
int
v;
int
count;
Quer(
int
vv,
int
c): v(vv),count(c) {}
};
vector
<
Edge
>
Graph[MAX];
vector
<
Quer
>
query[MAX];
bool
vis[MAX];
int
fa[MAX],n,q;
long
long
dist[MAX],ans[MAX];
void
init()
{
int
i;
for
(i
=
0
;i
<=
n;
++
i)
{
Graph[i].clear();
dist[i]
=
0
;
fa[i]
=
i;
vis[i]
=
false
;
}
for
(i
=
0
;i
<=
MAX;
++
i) query[i].clear();
}
int
Findset(
int
x)
{
while
(fa[x]
!=
x)
{
x
=
fa[x];
}
return
x;
}
void
Tarjan(
int
now,
long
long
cur_dis)
{
int
i,end,dis,v,count;
vis[now]
=
true
; dist[now]
=
cur_dis; fa[now]
=
now;
for
(i
=
0
;i
<
query[now].size();
++
i)
{
v
=
query[now][i].v; count
=
query[now][i].count;
if
(vis[v])
{
ans[count]
=
dist[now]
+
dist[v]
-
2
*
dist[Findset(v)];
}
}
for
(i
=
0
;i
<
Graph[now].size();
++
i)
{
end
=
Graph[now][i].end; dis
=
Graph[now][i].dis;
if
(
!
vis[end])
{
Tarjan(end,cur_dis
+
dis);
fa[end]
=
now;
}
}
}
int
main()
{
int
i,u,v,d;
while
(scanf(
"
%d
"
,
&
n)
&&
n)
{
init();
for
(i
=
1
;i
<
n;
++
i)
{
scanf(
"
%d %d
"
,
&
v,
&
d);
Graph[v].push_back(Edge(i,d));
Graph[i].push_back(Edge(v,d));
}
scanf(
"
%d
"
,
&
q);
for
(i
=
0
;i
<
q;
++
i)
{
scanf(
"
%d %d
"
,
&
u,
&
v);
query[u].push_back(Quer(v,i));
query[v].push_back(Quer(u,i));
}
Tarjan(
0
,
0
);
for
(i
=
0
;i
<
q;
++
i)
{
if
(i) printf(
"
"
);
cout
<<
ans[i];
}
printf(
"
\n
"
);
}
return
0
;
}