uva 103 - Stacking Boxes
题目读了好久总算看懂了，具有n维属性的箱子a和b，如果a fits b 仅当存在a的属性排列，使与b对应
的属性满足q(a)<q(b)，算法中定义二维数组，标记a是否fits b，然后深搜，当然得先把每个箱子的属性排序。
1
#include
<
stdio.h
>
2
#include
<
string
.h
>
3
#include
<
stdlib.h
>
4
5
struct
Box {
6
int
mt[
15
];
7
}bx[
40
];
8
bool
flg[
40
][
40
], f0[
40
];
9
int
rst[
40
], tp[
40
];
10
int
nb, mt;
11
int
max;
12
13
int
comp(
const
void
*
a,
const
void
*
b)
14
{
15
return
*
((
int
*
)a)
-
*
((
int
*
)b);
16
}
17
18
bool
fits(
int
i,
int
j)
19
{
20
for
(
int
k
=
0
; k
<
mt; k
++
)
21
if
(bx[i].mt[k]
>=
bx[j].mt[k])
return
false
;
22
return
true
;
23
}
24
25
void
DFS(
int
i,
int
t)
26
{
27
int
j;
28
for
(j
=
1
; j
<=
nb; j
++
) {
29
if
(flg[i][j]) {
30
tp[t
+
1
]
=
j;
31
DFS(j, t
+
1
);
32
}
33
}
34
if
(max
<
t) {
35
for
(j
=
1
; j
<=
t; j
++
)
36
rst[j]
=
tp[j];
37
max
=
t;
38
}
39
}
40
41
int
main()
42
{
43
int
i, j;
44
while
(scanf(
"
%d%d
"
,
&
nb,
&
mt)
!=
EOF) {
45
for
(i
=
1
; i
<=
nb; i
++
) {
46
for
(j
=
0
; j
<
mt; j
++
)
47
scanf(
"
%d
"
,
&
bx[i].mt[j]);
48
qsort(bx[i].mt, mt,
sizeof
(bx[i].mt[
0
]), comp);
49
}
50
memset(flg,
0
,
sizeof
(flg));
51
for
(i
=
1
; i
<=
nb; i
++
) {
52
for
(j
=
1
; j
<=
nb; j
++
) {
53
if
(i
==
j)
continue
;
54
if
(fits(i, j)) {
55
flg[i][j]
=
1
;
56
}
57
}
58
}
59
max
=
0
;
60
61
for
(i
=
1
; i
<=
nb; i
++
) {
62
tp[
1
]
=
i;
63
memset(f0,
0
,
sizeof
(f0));
64
f0[i]
=
1
;
65
DFS(i,
1
);
66
}
67
//
if a fits b , b fits c then a fits c
68
//
因此rst[]保存的就是期望的顺序
69
printf(
"
%d\n
"
, max);
70
71
for
(i
=
1
; i
<=
max; i
++
) {
72
printf(
"
%d
"
, rst[i]);
73
if
(i
!=
max) printf(
"
"
);
74
}
75
printf(
"
\n
"
);
76
}
77
return
0
;
78
}
79