USACO 2.1 Sorting A Three-Valued Sequence
Type: tricky
题目描述：
Sorting is one of the most frequently performed computational tasks. Consider the special sorting problem in which the records to be sorted have at most three different key values. This happens for instance when we sort medalists of a competition according to medal value, that is, gold medalists come first, followed by silver, and bronze medalists come last.
In this task the possible key values are the integers 1, 2 and 3. The required sorting order is non-decreasing. However, sorting has to be accomplished by a sequence of exchange operations. An exchange operation, defined by two position numbers p and q, exchanges the elements in positions p and q.
You are given a sequence of key values. Write a program that computes the minimal number of exchange operations that are necessary to make the sequence sorted.
INPUT FORMAT
| Line 1: | N (1 <= N <= 1000), the number of records to be sorted |
| Lines 2-N+1: | A single integer from the set {1, 2, 3} |
SAMPLE INPUT (file sort3.in)
9221333231
OUTPUT FORMAT
A single line containing the number of exchanges required
SAMPLE OUTPUT (file sort3.out)
4
思路：
对输入数列的两个copy扫描三遍；
第一遍，找出相同的，标记；
第二遍，对于不同的，找出能跟它配对的，比如：1->2, 2->1。 cnt ++;
第三遍，找出还没有标记过的，剩下的为三元对，1->2, 2->3, 3->1。 cnt +=2；
代码
/*
ID: superbi1
LANG: C
TASK: sort3
*/
#include
<
stdio.h
>
#include
<
string
.h
>
#include
<
stdlib.h
>
#define
NL 1001
struct
A {
int
t;
};
struct
A a1[NL], a2[NL];
int
flg[NL];
int
cmp(
const
void
*
a,
const
void
*
b)
{
return
((
struct
A
*
)a)
->
t
-
((
struct
A
*
)b)
->
t;
}
int
main()
{
FILE
*
in
=
fopen(
"
sort3.in
"
,
"
r
"
);
FILE
*
out
=
fopen(
"
sort3.out
"
,
"
w
"
);
int
n, i, j;
int
cnt
=
0
, t;
fscanf(
in
,
"
%d
"
,
&
n);
for
(i
=
0
; i
<
n; i
++
) {
fscanf(
in
,
"
%d
"
,
&
a1[i].t);
a2[i].t
=
a1[i].t;
}
qsort(a1, n,
sizeof
(a1[
0
]), cmp);
memset(flg,
0
,
sizeof
(flg));
for
(i
=
0
; i
<
n; i
++
) {
if
(a1[i].t
==
a2[i].t) flg[i]
=
1
;
}
for
(i
=
0
; i
<
n; i
++
) {
if
(
!
flg[i]) {
for
(j
=
i
+
1
; j
<
n; j
++
) {
if
(
!
flg[j]
&&
a1[i].t
==
a2[j].t
&&
a1[j].t
==
a2[i].t) {
cnt
++
;
flg[i]
=
1
;
flg[j]
=
1
;
break
;
}
}
}
}
t
=
0
;
for
(i
=
0
; i
<
n; i
++
) {
if
(
!
flg[i]) { printf(
"
%d %d\n
"
, a1[i].t, a2[i].t), t
++
; }
}
if
(t
>
0
) cnt
+=
t
/
3
*
2
;
fprintf(
out
,
"
%d\n
"
, cnt);
return
0
;
}