树形DP 题目库
树形DP，是一个很好理解的DP，树形结构很容易理解DP的！
在存储结构上，学习了左孩子，右兄弟的结构：
定义代码：
定义代码
struct
node{
int
val;
int
father;
int
child;
int
brother;
void
init()
{
father
=
child
=
brother
=
-
1
;
val
=
0
;
}
}tree[maxn];
添加边：
void
AddEdge(
int
a,
int
b,
int
c)
{
tree[b].father
=
a;
tree[b].brother
=
tree[a].child;
tree[a].child
=
b;
tree[b].val
=
c;
//
把边的值赋给了某点
}
hdu 1520 Anniversary party
http://acm.hdu.edu.cn/showproblem.php?pid=1520
简单树形DP，做的第一个DP。
代码
/*
简单树形DP题，比较好的地方是用到了左孩子，右兄弟的森林存储结构
*/
#include
<
iostream
>
using
namespace
std;
const
long
maxn
=
6005
;
#define
max(a,b) (a>b?a:b)
int
n;
struct
node
{
int
Take;
int
Not;
int
father;
int
child;
int
brother;
void
init()
{
father
=
child
=
brother
=
-
1
;
Take
=
0
;
Not
=
0
;
}
int
Max()
{
return
Take
>
Not
?
Take:Not;
}
}tree[maxn];
void
Init()
{
int
i,l,k;
for
(i
=
1
;i
<=
n;i
++
)
{
tree[i].init();
scanf(
"
%d
"
,
&
tree[i].Take);
}
while
(scanf(
"
%d%d
"
,
&
l,
&
k)
!=
EOF
&&
(l
!=
0
||
k
!=
0
))
{
tree[l].father
=
k;
tree[l].brother
=
tree[k].child;
tree[k].child
=
l;
}
//
把森林合成一棵树
int
pre
=
-
1
;
for
(i
=
1
;i
<=
n;i
++
)
{
if
(tree[i].father
==
-
1
)
{
tree[i].father
=
0
;
tree[i].brother
=
pre;
tree[
0
].child
=
i;
pre
=
i;
}
}
tree[
0
].father
=
-
1
;
tree[
0
].brother
=
-
1
;
tree[
0
].Take
=
0
;
tree[
0
].Not
=
0
;
}
void
Sum(
int
dir)
{
int
c
=
tree[dir].child;
while
(c
!=
-
1
)
{
Sum(c);
tree[dir].Take
+=
tree[c].Not;
tree[dir].Not
+=
tree[c].Max();
c
=
tree[c].brother;
}
}
void
print()
{
int
num
=
tree[
0
].Max();
printf(
"
%d\n
"
,num);
}
int
main()
{
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
Init();
Sum(
0
);
print();
}
return
0
;
}
hdu 3660 Alice and Bob's Trip
http://acm.hdu.edu.cn/showproblem.php?pid=3660
Bob选最大子结点走，Alice选最小子结点走！
但因为要符合《L，R》，所以加一个判断条件：
tree[dir].sum + tree[c].ans + tree[c].val >= L && tree[dir].sum + tree[c].ans + tree[c].val <= R
//c是dir点的一个子结点，sum表示从0结点到该点的距离，ans表示最优子结点的距离
代码
#include
<
iostream
>
using
namespace
std;
const
long
inf
=
1000000000
;
const
long
maxn
=
500005
;
#define
max(a,b) (a>b?a:b)
#define
min(a,b) (a>b?b:a)
int
n,L,R;
struct
node
{
int
child;
int
father;
int
brother;
int
val;
int
sum;//从0结点到该点的值
int
ans;//子结点到该点的最优值
void
Init()
{
child
=
father
=
brother
=
-
1
;
val
=
0
;sum
=
0
;ans
=
0
;
}
}tree[maxn];
void
Init()
{
int
i;
int
a,b,c;
for
(i
=
0
;i
<
n;i
++
)
tree[i].Init();
for
(i
=
0
;i
<
n
-
1
;i
++
)
{
scanf(
"
%d%d%d
"
,
&
a,
&
b,
&
c);
tree[b].father
=
a;
tree[b].brother
=
tree[a].child;
tree[a].child
=
b;
tree[b].val
=
c;
}
}
void
Sum(
int
dir)
{
int
c
=
tree[dir].child;
while
(c
!=
-
1
)
{
tree[c].sum
=
tree[tree[c].father].sum
+
tree[c].val;
Sum(c);
c
=
tree[c].brother;
}
}
void
Bfs(
int
dir,
bool
bol)
{
if
(bol
==
0
)
tree[dir].ans
=
0
;
else
tree[dir].ans
=
inf;
int
c
=
tree[dir].child;
if
(c
==
-
1
)
tree[dir].ans
=
0
;
while
(c
!=
-
1
)
{
Bfs(c,
!
bol);
if
(tree[dir].sum
+
tree[c].ans
+
tree[c].val
>=
L
&&
tree[dir].sum
+
tree[c].ans
+
tree[c].val
<=
R)
{
if
(bol
==
0
)
{
tree[dir].ans
=
max(tree[dir].ans,tree[c].ans
+
tree[c].val);
}
else
{
tree[dir].ans
=
min(tree[dir].ans,tree[c].ans
+
tree[c].val);
}
}
c
=
tree[c].brother;
}
}
void
Print()
{
if
(tree[
0
].ans
>=
L
&&
tree[
0
].ans
<=
R)
printf(
"
%d\n
"
,tree[
0
].ans);
else
printf(
"
Oh, my god!\n
"
);
}
int
main()
{
while
(scanf(
"
%d%d%d
"
,
&
n,
&
L,
&
R)
!=
EOF)
{
Init();
Sum(
0
);
Bfs(
0
,
0
);
Print();
}
return
0
;
}