TJU 3758. Jewel 【划分树，树状数组】
http://acm.tju.edu.cn/toj/showp3758.html
TJU 3758. Jewel
模拟题，开始数组为空，有以下操作：
Insert x
Put a bead with size x to the right of the chain (0 < x < 231, and x is different from all the sizes of beads currently in the chain)
Query_1 s t k
Query the k-th smallest bead between position s and t, inclusive. You can assume 1 ≤ s ≤ t ≤ L, (L is the length of the current chain), and 1 ≤ k ≤ min (100, t - s + 1)
Query_2 x
Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 ≤ k ≤ L, L is the length of the current chain)
要求统计每种问题的答案
分析：对于问题1和3可用划分树解决，对于问题2用树状数组解决，注意用树状数组前需要对数据离散化
View Code
#include
<
stdio.h
>
#include
<
algorithm
>
#include
<
iostream
>
using
namespace
std;
typedef
long
long
LL;
struct
Q
//
问题
{
int
kind;
//
问题类型
int
x,k;
int
s,t;
int
ans;
}q[
35000
*
4
+
10
];
const
int
MAXN
=
100000
+
10
;
int
tree[
22
][MAXN],toleft[
22
][MAXN];
int
sorted[MAXN];
//
已经排好序的数据
int
c[MAXN];
inline
int
low(
int
x)
{
return
(x
&
(
-
x));
}
void
add(
int
x,
int
n)
{
while
( x
<=
n)
{
c[x]
++
;
x
+=
low(x);
}
}
LL Sum(
int
x)
{
LL ans
=
0
;
while
(x
>
0
)
{
ans
+=
c[x];
x
-=
low(x);
}
return
ans;
}
void
build_tree(
int
left,
int
right,
int
deep)
//
建树
{
if
(left
==
right)
return
;
int
mid
=
(left
+
right)
>>
1
;
int
i;
int
same
=
mid
-
left
+
1
;
//
位于左子树的数据
for
(i
=
left;i
<=
right;i
++
)
//
计算放于左子树中与中位数相等的数字个数
{
if
(tree[deep][i]
<
sorted[mid])
same
--
;
}
int
ls
=
left;
int
rs
=
mid
+
1
;
for
(i
=
left;i
<=
right;i
++
)
{
int
flag
=
0
;
if
((tree[deep][i]
<
sorted[mid])
||
(tree[deep][i]
==
sorted[mid]
&&
same
>
0
))
{
flag
=
1
;
tree[deep
+
1
][ls
++
]
=
tree[deep][i];
if
(tree[deep][i]
==
sorted[mid])same
--
;
}
else
{
tree[deep
+
1
][rs
++
]
=
tree[deep][i];
}
toleft[deep][i]
=
toleft[deep][i
-
1
]
+
flag;
}
build_tree(left,mid,deep
+
1
);
build_tree(mid
+
1
,right,deep
+
1
);
}
int
query(
int
left,
int
right,
int
k,
int
L,
int
R,
int
deep)
{
if
(left
==
right)
return
tree[deep][left];
int
mid
=
(L
+
R)
>>
1
;
int
x
=
toleft[deep][left
-
1
]
-
toleft[deep][L
-
1
];
//
位于left左边的放于左子树中的数字个数
int
y
=
toleft[deep][right]
-
toleft[deep][L
-
1
];
//
到right为止位于左子树的个数
int
ry
=
right
-
L
-
y;
//
到right右边为止位于右子树的数字个数
int
cnt
=
y
-
x;
//
[left,right]区间内放到左子树中的个数
int
rx
=
left
-
L
-
x;
//
left左边放在右子树中的数字个数
if
(cnt
>=
k)
return
query(L
+
x,L
+
y
-
1
,k,L,mid,deep
+
1
);
else
return
query(mid
+
rx
+
1
,mid
+
1
+
ry,k
-
cnt,mid
+
1
,R,deep
+
1
);
}
int
find(
int
x,
int
left,
int
right)
{
int
ans
=
0
;
while
(left
<=
right)
{
int
mid
=
(left
+
right)
>>
1
;
if
(x
==
sorted[mid])
return
mid;
if
(x
<
sorted[mid])
right
=
mid
-
1
;
else
left
=
mid
+
1
;
}
return
ans;
}
int
main()
{
int
op
=
0
;
int
cases
=
0
;
while
(scanf(
"
%d
"
,
&
op)
!=
EOF)
{
cases
++
;
int
n
=
1
;
int
qnum
=
0
;
//
问题个数
while
(op
--
)
{
char
str[
20
];
scanf(
"
%s
"
,str);
if
(str[
0
]
==
'
I
'
)
{
scanf(
"
%d
"
,
&
sorted[n]);
tree[
0
][n]
=
sorted[n];
n
++
;
}
else
{
int
len
=
strlen(str);
if
(str[len
-
1
]
==
'
1
'
)
{
q[qnum].kind
=
1
;
scanf(
"
%d%d%d
"
,
&
q[qnum].s,
&
q[qnum].t,
&
q[qnum].k);
qnum
++
;
}
else
if
(str[len
-
1
]
==
'
2
'
)
{
q[qnum].kind
=
2
;
scanf(
"
%d
"
,
&
q[qnum].x);
q[qnum].t
=
n
-
1
;
qnum
++
;
}
else
if
(str[len
-
1
]
==
'
3
'
)
{
q[qnum].kind
=
3
;
scanf(
"
%d
"
,
&
q[qnum].k);
q[qnum].s
=
1
;
q[qnum].t
=
n
-
1
;
qnum
++
;
}
}
}
sort(sorted
+
1
,sorted
+
n);
memset(c,
0
,
sizeof
(c));
int
i,pre
=
1
;
build_tree(
1
,n
-
1
,
0
);
LL sum1
=
0
,sum2
=
0
,sum3
=
0
;
for
(i
=
0
;i
<
qnum;i
++
)
{
if
(q[i].kind
==
1
)
{
//
query(int left,int right,int k,int L,int R,int deep)
sum1
+=
query(q[i].s,q[i].t,q[i].k,
1
,n
-
1
,
0
);
}
else
if
(q[i].kind
==
3
)
{
sum3
+=
query(q[i].s,q[i].t,q[i].k,
1
,n
-
1
,
0
);
}
else
{
for
(;pre
<=
q[i].t;pre
++
)
{
int
index
=
find(tree[
0
][pre],
1
,n
-
1
);
add(index,n
-
1
);
}
int
index
=
find(q[i].x,
1
,n
-
1
);
//
printf("%ld\n",Sum(index));
sum2
+=
Sum(index);
}
}
printf(
"
Case %d:\n
"
,cases);
cout
<<
sum1
<<
endl;
cout
<<
sum2
<<
endl;
cout
<<
sum3
<<
endl;
//
printf("%I64d\n%I64d\n%I64d\n",sum1,sum2,sum3);
}
return
0
;
}