HDU 3635 Dragon Balls [并查集]
题意:有n颗龙珠分别顺序放在n个城市中，有两种操作
（1）T A B：把A所在城市的所有龙珠转运到B所在的城市。
（2）Q A ：查询A所在的城市，所在城市的龙珠个数，A被转运了几次
思路：开始就是想用并查集做，但是处理细节很麻烦，每个节点记录被转运的次数，路径压缩时就可以修改节点，别的方法没想到...输入数据很多cin回超时。
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
bitset
>
#include
<
queue
>
#include
<
vector
>
#include
<
map
>
#include
<
stack
>
#include
<
set
>
#include
<
list
>
#include
<
deque
>
#include
<
cstdlib
>
#include
<
cstring
>
const
int
BORDER
=
(
1
<<
20
)
-
1
;
const
int
MAXSIZE
=
37
;
const
int
MAXN
=
20500
;
const
int
INF
=
0x3f3f3f3f
;
#define
CLR(x,y) memset(x,y,sizeof(x))
#define
MIN(m,v) (m)<(v)?(m):(v)
#define
MAX(m,v) (m)>(v)?(m):(v)
#define
ABS(x) ((x)>0?(x):-(x))
#define
REP(i,x,y) for(i=x;i<y;++i)
#define
getbit(a,i) (a&(1<<i))
#define
setbit(a,i) (a|(1<<i))
using
namespace
std;
int
pre[MAXN],ct[MAXN],num[MAXN];
int
ic,n,m;
int
find_set(
int
x)
{
int
root
=
x;
int
tmp,tmp_c,t;
t
=
0
;
while
(pre[root]
>=
0
)
{
root
=
pre[root];
t
=
t
+
ct[root];
}
ct[x]
+=
t;
while
(x
!=
root)
{
tmp
=
pre[x];
tmp_c
=
ct[tmp];
pre[x]
=
root;
ct[tmp]
=
t;
t
-=
tmp_c;
x
=
tmp;
}
return
root;
}
void
union_set(
const
int
&
root1,
const
int
&
root2)
{
if
(num[root1]
==
0
)
return
;
num[root2]
+=
num[root1];
num[root1]
=
0
;
++
ct[root1];
pre[root2]
+=
pre[root1];
pre[root1]
=
root2;
return
;
}
int
init()
{
CLR(ct,
0
);
CLR(pre,
-
1
);
return
0
;
}
int
input()
{
scanf(
"
%d%d
"
,
&
n,
&
m);
int
i;
REP(i,
0
,n
+
1
)
num[i]
=
1
;
return
0
;
}
int
work()
{
int
i,a,b;
int
pos,cnt,t;
int
root1,root2;
char
ch[
3
];
printf(
"
Case %d:\n
"
,ic);
++
ic;
REP(i,
0
,m)
{
scanf(
"
%s
"
,ch);
if
(ch[
0
]
==
'
T
'
)
{
scanf(
"
%d %d
"
,
&
a,
&
b);
root1
=
find_set(a);
root2
=
find_set(b);
union_set(root1,root2);
}
else
{
scanf(
"
%d
"
,
&
a);
pos
=
find_set(a);
printf(
"
%d %d %d\n
"
,pos,num[pos],ct[a]);
}
}
return
0
;
}
int
main()
{
ic
=
1
;
int
tt;
scanf(
"
%d
"
,
&
tt);
while
(tt
--
)
{
init();
input();
work();
}
return
0
;
}