USACO 1.4 Packing Rectangles
这个题好久以前就看过了，但当时没思路，现在重新看，发现还是可以做的。
思路：
根据题目给出的六种layout，对每种的每个位置有4中选择（即4个矩形)，另外4个矩形旋转后又可形成4个矩形。
基本上是枚举所有的情况，然后找最小的面积值。最后一种layout，其实又要分四种情况的。
代码
1
/*
2
ID: superbi1
3
LANG: C
4
TASK: packrec
5
*/
6
#include
<
stdio.h
>
7
#include
<
string
.h
>
8
#include
<
stdlib.h
>
9
#define
MAX(x,y) ((x)>(y)?(x):(y))
10
#define
SWP(x,y,t) ((t)=(x), (x)=(y), (y)=(t))
11
12
typedef
struct
Rct {
13
int
area, a, b;
14
}Rct;
15
Rct rct[
3000
];
16
int
nrt;
17
int
r[
9
][
2
];
18
19
void
calt(
int
t)
20
{
21
int
I, K1, K2, K3, K4, tmp;
22
char
flg[
9
];
23
for
(I
=
1
; I
<=
8
; I
++
) flg[I]
=
0
;
24
for
(K1
=
1
; K1
<=
8
; K1
++
) {
25
flg[K1]
=
1
;
26
for
(K2
=
1
; K2
<=
8
; K2
++
) {
27
if
(flg[K2]
||
flg[(K2
+
4
)
%
8
==
0
?
8
:(K2
+
4
)
%
8
])
continue
;
28
flg[K2]
=
1
;
29
for
(K3
=
1
; K3
<=
8
; K3
++
) {
30
if
(flg[K3]
||
flg[(K3
+
4
)
%
8
==
0
?
8
:(K3
+
4
)
%
8
])
continue
;
31
flg[K3]
=
1
;
32
for
(K4
=
1
; K4
<=
8
; K4
++
) {
33
if
(flg[K4]
||
flg[(K4
+
4
)
%
8
==
0
?
8
:(K4
+
4
)
%
8
])
continue
;
34
switch
(t) {
35
case
1
:
36
rct[nrt].b
=
r[K1][
1
]
+
r[K2][
1
]
+
r[K3][
1
]
+
r[K4][
1
];
37
rct[nrt].a
=
MAX(r[K1][
0
], MAX(r[K2][
0
], MAX(r[K3][
0
], r[K4][
0
])));
38
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
39
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
40
nrt
++
;
41
break
;
42
case
2
:
43
rct[nrt].b
=
MAX(r[K4][
1
], r[K1][
1
]
+
r[K2][
1
]
+
r[K3][
1
]);
44
rct[nrt].a
=
MAX(r[K1][
0
], MAX(r[K2][
0
], r[K3][
0
]))
+
r[K4][
0
];
45
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
46
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
47
nrt
++
;
48
break
;
49
case
3
:
50
rct[nrt].b
=
MAX(r[K1][
1
]
+
r[K2][
1
]
+
r[K3][
1
], r[K4][
1
]
+
r[K3][
1
]);
51
rct[nrt].a
=
MAX(MAX(r[K1][
0
], r[K2][
0
])
+
r[K4][
0
], r[K3][
0
]);
52
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
53
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
54
nrt
++
;
55
break
;
56
case
4
:
57
rct[nrt].b
=
MAX(r[K2][
1
], r[K4][
1
])
+
r[K1][
1
]
+
r[K3][
1
];
58
rct[nrt].a
=
MAX(r[K1][
0
], MAX(r[K3][
0
], r[K2][
0
]
+
r[K4][
0
]));
59
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
60
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
61
nrt
++
;
62
break
;
63
case
5
:
64
rct[nrt].b
=
MAX(r[K1][
1
], r[K2][
1
])
+
r[K3][
1
]
+
r[K4][
1
];
65
rct[nrt].a
=
MAX(r[K1][
0
]
+
r[K2][
0
], MAX(r[K3][
0
], r[K4][
0
]));
66
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
67
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
68
nrt
++
;
69
break
;
70
case
6
:
71
if
(r[K1][
1
]
>
r[K2][
1
]) {
72
if
(r[K2][
0
]
>
r[K3][
0
]) {
73
rct[nrt].a
=
MAX(r[K1][
0
]
+
r[K2][
0
], r[K3][
0
]
+
r[K4][
0
]);
74
}
else
{
75
rct[nrt].a
=
MAX(r[K1][
0
], r[K4][
0
])
+
r[K3][
0
];
76
}
77
}
else
{
78
if
(r[K2][
0
]
<
r[K3][
0
]) {
79
rct[nrt].a
=
MAX(r[K1][
0
]
+
r[K2][
0
], r[K3][
0
]
+
r[K4][
0
]);
80
}
else
{
81
rct[nrt].a
=
MAX(r[K1][
0
], r[K4][
0
])
+
r[K2][
0
];
82
}
83
}
84
rct[nrt].b
=
MAX(r[K1][
1
]
+
r[K4][
1
], r[K2][
1
]
+
r[K3][
1
]);
85
if
(rct[nrt].a
>
rct[nrt].b) SWP(rct[nrt].a, rct[nrt].b, tmp);
86
rct[nrt].area
=
rct[nrt].a
*
rct[nrt].b;
87
nrt
++
;
88
break
;
89
default
:
90
break
;
91
}
92
}
93
flg[K3]
=
0
;
94
}
95
flg[K2]
=
0
;
96
}
97
flg[K1]
=
0
;
98
}
99
}
100
101
int
cmp(
const
void
*
a,
const
void
*
b)
102
{
103
if
(((Rct
*
)a)
->
area
!=
((Rct
*
)b)
->
area)
104
return
((Rct
*
)a)
->
area
-
((Rct
*
)b)
->
area;
105
return
((Rct
*
)a)
->
a
-
((Rct
*
)b)
->
a;
106
}
107
108
int
main()
109
{
110
int
I, min, a;
111
FILE
*
fin
=
fopen(
"
packrec.in
"
,
"
r
"
);
112
FILE
*
fout
=
fopen(
"
packrec.out
"
,
"
w
"
);
113
for
(I
=
1
; I
<=
4
; I
++
) {
114
fscanf(fin,
"
%d%d
"
,
&
r[I][
0
],
&
r[I][
1
]);
115
r[I
+
4
][
0
]
=
r[I][
1
];
116
r[I
+
4
][
1
]
=
r[I][
0
];
117
}
118
nrt
=
0
;
119
for
(I
=
1
; I
<=
6
; I
++
) {
120
calt(I);
121
}
122
qsort(rct, nrt,
sizeof
(rct[
0
]), cmp);
123
min
=
rct[
0
].area;
124
a
=
rct[
0
].a;
125
fprintf(fout,
"
%d\n%d %d\n
"
, min, rct[
0
].a, rct[
0
].b);
126
for
(I
=
1
; I
<
nrt; I
++
) {
127
if
(rct[I].area
==
min
&&
rct[I].a
!=
a) {
128
a
=
rct[I].a;
129
fprintf(fout,
"
%d %d\n
"
, rct[I].a, rct[I].b);
130
}
else
if
(rct[I].area
>
min)
break
;
131
}
132
return
0
;
133
}