LCA与RMQ
参见topcoder的算法教程，Range Minimum Query and Lowest Common Ancestor
JOJ 一个题目：
2408 Beautiful girl， 题意：一颗树，给定三个顶点A、B、C，判断A与B之间的路径是否可以经过C
代码：
代码
/*
* LCA->RMQ
*/
#include
<
cstdio
>
#include
<
vector
>
#include
<
cmath
>
using
namespace
std;
#define
MY_MAX 50010
vector
<
int
>
graph[MY_MAX];
void
read(
int
n){
int
i, u, v;
for
(i
=
0
; i
<
n;
++
i)
if
(
!
graph[i].empty())
graph[i].clear();
for
(i
=
1
; i
<
n;
++
i){
scanf(
"
%d %d
"
,
&
u,
&
v);
graph[u].push_back(v);
graph[v].push_back(u);
}
}
int
E[MY_MAX
*
2
];
//
E[i] is the label of i-th visited node in the tour
int
L[MY_MAX
*
2
];
//
L[i] is the level of node E[i]
int
H[MY_MAX];
//
the index of the first occurrence of node i in E
bool
visited[MY_MAX];
void
DFS(
int
u,
int
d,
int
&
k){
visited[u]
=
true
;
E[k]
=
u;
L[k]
=
d;
H[u]
=
k;
++
k;
int
v;
for
(size_t i
=
0
; i
<
graph[u].size();
++
i){
v
=
graph[u][i];
if
(
!
visited[v]){
DFS(v, d
+
1
, k);
E[k]
=
u;
L[k]
=
d;
++
k;
}
}
}
int
M[MY_MAX
*
2
][
20
];
//
存储的索引
void
preprocess(
int
a[],
int
n){
int
i, j;
//
initialize for the intervals with length 1
for
(i
=
0
; i
<
n;
++
i)
M[i][
0
]
=
i;
//
compute values from smaller to bigger intervals
for
(j
=
1
;
1
<<
j
<=
n;
++
j){
for
(i
=
0
; i
+
(
1
<<
j)
-
1
<
n;
++
i)
if
(a[M[i][j
-
1
]]
<
a[M[i
+
(
1
<<
(j
-
1
))][j
-
1
]])
M[i][j]
=
M[i][j
-
1
];
else
M[i][j]
=
M[i
+
(
1
<<
(j
-
1
))][j
-
1
];
}
}
int
RMQ(
int
a[],
int
i,
int
j){
if
(i
>
j)
swap(i, j);
int
k;
k
=
int
(log(j
-
i
+
1
)
/
log(
2.0
));
if
(a[M[i][k]]
<
a[M[j
-
(
1
<<
k)
+
1
][k]])
return
M[i][k];
else
return
M[j
-
(
1
<<
k)
+
1
][k];
}
int
main(){
//
freopen("in", "r", stdin);
int
n, m, k, case_num
=
1
;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF){
read(n);
for
(k
=
0
; k
<
n;
++
k)
visited[k]
=
false
;
k
=
0
;
DFS(
0
,
0
, k);
preprocess(L,
2
*
n
-
1
);
int
a, b, c;
int
lab, lac, lbc;
if
(case_num
!=
1
)
printf(
"
\n
"
);
printf(
"
Case %d:\n
"
, case_num
++
);
scanf(
"
%d
"
,
&
m);
for
(k
=
0
; k
<
m;
++
k){
scanf(
"
%d %d %d
"
,
&
a,
&
b,
&
c);
lab
=
E[RMQ(L, H[a], H[b])];
lac
=
E[RMQ(L, H[a], H[c])];
lbc
=
E[RMQ(L, H[b], H[c])];
if
(a
==
c
||
b
==
c
||
(lab
==
lac
&&
lbc
==
c)
||
(lab
==
lbc
&&
lac
==
c))
printf(
"
Yes\n
"
);
else
printf(
"
No\n
"
);
}
}
}