zafu 1467
终于把校赛最后一题过了，二分的地方写的很差，以后注意改进。思路就是从前向后求LIS最长递增子序列，以a[i]为底LIS值记录在dp1[i],
然后从后向前求LDS最长递减子序列，以a[i]为底LDS值记录在dp2[i]
则如果 dp1[i] + dp2[i]== dp[n]+1( 即最长长度加1 ）,则说明a[i]是最长中的某个数，如果这些符合的dp1[i]值只出现一次，则其为
Key Number
题目链接 http://info.zjfc.edu.cn/acm/problemDetail.aspx?pid=1467
代码
#include
<
stdio.h
>
#define
maxn 100005
int
dp1[maxn],dp2[maxn];
int
a[maxn];
int
hash[maxn];
int
stack[maxn];
int
n;
int
kin;
void
LIS(
int
k )
{
int
i,j;
int
end,front,mid;
int
top
=
1
;
stack[top]
=
a[k];
dp1[
1
]
=
1
;
for
( i
=
k
+
1
; i
<=
n ; i
++
)
{
if
( a[i]
>
stack[ top ] )
{
stack[
++
top ]
=
a[i] ;
dp1[i]
=
top;
//
}
else
if
( a[i]
<=
stack[
1
] )
{
stack[
1
]
=
a[i];
dp1[i]
=
1
;
}
else
{
front
=
1
;
end
=
top;
while
( end
>=
front )
{
mid
=
( front
+
end )
/
2
;
if
( a[i]
==
stack[mid] )
{
dp1[i]
=
mid;
break
;
}
else
if
( a[i]
>
stack[mid]
&&
a[i]
<=
stack[mid
+
1
] )
{
stack[mid
+
1
]
=
a[i] ;
dp1[i]
=
mid
+
1
;
//
break
;
}
else
if
( a[i]
<
stack[mid] )
end
=
mid
-
1
;
else
if
(a[i]
>
stack[mid
+
1
] )
front
=
mid
+
1
;
}
}
}
kin
=
top;
}
void
LDS(
int
k )
{
int
i,j;
int
end,front,mid;
int
top
=
1
;
stack[top]
=
a[k];
dp2[k]
=
1
;
for
( i
=
k
-
1
; i
>
0
; i
--
)
{
if
( a[i]
<
stack[ top ] )
{
stack[
++
top ]
=
a[i] ;
dp2[i]
=
top;
}
else
if
( a[i]
>=
stack[
1
] )
{
stack[
1
]
=
a[i];
dp2[i]
=
1
;
}
else
{
front
=
1
;
end
=
top;
while
( end
>=
front )
{
mid
=
( front
+
end )
/
2
;
if
( a[i]
==
stack[mid] )
{
dp2[i]
=
mid;
break
;
}
else
if
( a[i]
<
stack[mid]
&&
a[i]
>=
stack[mid
+
1
] )
{
stack[mid
+
1
]
=
a[i] ;
dp2[i]
=
mid
+
1
;
break
;
}
else
if
( a[i]
>
stack[mid] )
end
=
mid
-
1
;
else
if
(a[i]
<
stack[mid
+
1
] )
front
=
mid
+
1
;
}
}
}
}
int
main()
{
int
t,i,j,k;
scanf(
"
%d
"
,
&
t);
while
( t
--
&&
scanf(
"
%d
"
,
&
n)
!=
EOF )
{
for
( i
=
0
; i
<=
n; i
++
)
//
初始
hash[ i ]
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
scanf(
"
%d
"
,
&
a[i] );
LIS(
1
) ;
LDS( n ) ;
for
( i
=
1
;i
<=
n;i
++
)
{
if
( dp1[i]
+
dp2[i]
==
kin
+
1
)
hash[ dp1[i] ]
++
;
}
int
sum
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
if
( hash[i]
==
1
)
sum
++
;
printf(
"
%d\n
"
,sum);
}
return
0
;
}