POJ 1655 Balancing Act (dfs)
题意：给定一棵树，问拿掉那个点，能使余下的各个子树结点个数上限最小。
看了半天，只知道是深搜，可是想了很久也没想出实现方法，各种菜~~~~看了大牛的实现，才恍然大悟........dfs的灵活之处自己需要时间去掌握啊.
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
set
>
#include
<
queue
>
#include
<
vector
>
using
namespace
std;
const
int
BORDER
=
(
1
<<
20
)
-
1
;
#define
MAXN 20020
#define
INF 0x7ffffff
#define
CLR(x,y) memset(x,y,sizeof(x))
#define
ADD(x) ((++x)&BORDER)
#define
IN(x) scanf("%d",&x)
#define
OUT(x) printf("%d\n",x)
#define
MIN(m,v) (m)<(v)?(m):(v)
#define
MAX(m,v) (m)>(v)?(m):(v)
#define
ABS(x) (x>0?x:-x)
struct
EDGE{
int
v,next;
}edge[MAXN
*
2
];
int
n,m,cnt,mmin_index,mmin_cnt,index;
int
net[MAXN],visit[MAXN];
void
add_edge(
const
int
&
u,
const
int
&
v)
{
edge[index].v
=
v;
edge[index].next
=
net[u];
net[u]
=
index;
++
index;
edge[index].v
=
u;
edge[index].next
=
net[v];
net[v]
=
index;
++
index;
}
int
init()
{
index
=
0
;
mmin_index
=
INF;
mmin_cnt
=
INF;
CLR(net,
-
1
);
CLR(visit,
0
);
return
0
;
}
int
dfs(
const
int
&
u)
{
int
i,v;
int
mmax
=
0
;
int
sum
=
0
;
visit[u]
=
1
;
for
(i
=
net[u]; i
!=
-
1
; i
=
edge[i].next)
{
v
=
edge[i].v;
if
(
!
visit[v])
{
int
tmp
=
dfs(v);
mmax
=
MAX(mmax,tmp);
sum
+=
tmp;
}
}
mmax
=
MAX(mmax,n
-
sum
-
1
);
if
(mmin_cnt
>=
mmax)
{
if
(mmin_cnt
==
mmax
&&
mmin_index
<
u)
;
else
{
mmin_cnt
=
mmax;
mmin_index
=
u;
}
}
return
sum
+
1
;
}
int
input()
{
int
i,j,tmp,a,b;
IN(n);
tmp
=
n
-
1
;
for
(i
=
0
; i
<
tmp;
++
i)
{
scanf(
"
%d %d
"
,
&
a,
&
b);
add_edge(a,b);
}
return
0
;
}
int
work()
{
dfs(
1
);
printf(
"
%d %d\n
"
,mmin_index,mmin_cnt);
return
0
;
}
int
main()
{
int
i,j,tt,tmp;
IN(tt);
while
(tt
--
)
{
init();
input();
work();
}
return
0
;
}