USACOTraining.Packing Rectangles
这道题确实是公认的USACO Tainning Section1中最BT的一道，题目大意是有4个矩形，问所需要的最小面积的盒子把这些矩形装起来。
学习别人Blog上的方法后过的。
就是枚举出所有的情况A(4, 4) * (2 ^ 4).
1
#include
<
iostream
>
2
#include
<
string
>
3
#include
<
algorithm
>
4
#include
<
string
.h
>
5
#include
<
vector
>
6
#include
<
math.h
>
7
#include
<
time.h
>
8
using
namespace
std;
9
10
const
int
INF
=
40005
;
11
12
struct
REC
13
{
14
int
w, h;
15
REC(){}
16
REC(
int
ww,
int
hh) : w(ww), h(hh)
17
{
18
if
(w
>
h) swap(w, h);
19
}
20
bool
operator
<
(
const
REC
&
t)
const
21
{
22
if
(w
!=
t.w)
return
w
<
t.w;
23
else
return
h
<
t.h;
24
}
25
}rec[
4
];
26
27
int
use[
4
];
28
int
repeat[
1005
][
1005
];
29
int
fmaxArea;
30
vector
<
REC
>
ans;
31
32
int
fmax(
int
a,
int
b) {
return
max(a, b);}
33
int
fmax(
int
a,
int
b,
int
c) {
return
max(max(a, b), c);}
34
int
fmax(
int
a,
int
b,
int
c,
int
d) {
return
max(max(a, b), max(c, d));}
35
36
void
kill(
int
w,
int
h)
37
{
38
if
(w
*
h
<
fmaxArea)
39
{
40
fmaxArea
=
w
*
h;
41
ans.clear();
42
ans.push_back(REC(w, h));
43
}
44
else
if
(w
*
h
==
fmaxArea)
45
{
46
ans.push_back(REC(w, h));
47
}
48
}
49
50
void
out
()
51
{
52
memset(repeat,
0
,
sizeof
(repeat));
53
sort(ans.begin(), ans.end());
54
55
printf(
"
%d\n
"
, fmaxArea);
56
for
(
int
i
=
0
; i
<
ans.size(); i
++
)
57
{
58
int
w
=
ans[i].w;
59
int
h
=
ans[i].h;
60
61
if
(repeat[w][h]
==
0
)
62
{
63
repeat[w][h]
=
1
;
64
printf(
"
%d %d\n
"
, w, h);
65
}
66
}
67
}
68
69
void
sol()
70
{
71
int
ww, hh;
72
int
w1
=
rec[use[
0
]].w;
73
int
w2
=
rec[use[
1
]].w;
74
int
w3
=
rec[use[
2
]].w;
75
int
w4
=
rec[use[
3
]].w;
76
77
int
h1
=
rec[use[
0
]].h;
78
int
h2
=
rec[use[
1
]].h;
79
int
h3
=
rec[use[
2
]].h;
80
int
h4
=
rec[use[
3
]].h;
81
82
//
1
83
ww
=
w1
+
w2
+
w3
+
w4;
84
hh
=
fmax(h1,h2,h3,h4);
85
kill(ww, hh);
86
87
//
2
88
ww
=
fmax(w1
+
w2
+
w3,w4);
89
hh
=
fmax(h1,h2,h3)
+
h4;
90
kill(ww, hh);
91
92
//
3
93
ww
=
fmax(w1
+
w2,w3)
+
w4;
94
hh
=
fmax(h1
+
h3,h2
+
h3,h4);
95
kill(ww, hh);
96
97
//
4
98
ww
=
w1
+
w2
+
fmax(w3,w4);
99
hh
=
fmax(h1,h3
+
h4,h2);
100
kill(ww, hh);
101
102
//
5
103
if
(h3
>=
h2
+
h4) ww
=
fmax(w1,w3
+
w2,w3
+
w4);
104
else
if
(h3
>
h4) ww
=
fmax(w1
+
w2,w2
+
w3,w3
+
w4);
105
else
if
(h3
==
h4) ww
=
fmax(w1
+
w2,w3
+
w4);
106
else
if
(h3
>
h4
-
h1) ww
=
fmax(w1
+
w2,w1
+
w4,w3
+
w4);
107
else
ww
=
fmax(w2,w1
+
w4,w3
+
w4);
108
109
hh
=
fmax(h1
+
h3,h2
+
h4);
110
kill(ww, hh);
111
}
112
113
void
rot(
int
x)
114
{
115
if
(x
==
4
)
116
{
117
sol();
118
return
;
119
}
120
rot(x
+
1
);
121
swap(rec[x].h, rec[x].w);
122
rot(x
+
1
);
123
swap(rec[x].h, rec[x].w);
124
}
125
126
void
dfs(
int
x)
127
{
128
if
(x
==
4
)
129
{
130
//
说明4个都选好位置放下
131
//
然后选择方向
132
rot(
0
);
133
return
;
134
}
135
136
for
(
int
i
=
0
; i
<
4
; i
++
)
137
{
138
if
(use[i]
==
-
1
)
139
{
140
use[i]
=
x;
141
dfs(x
+
1
);
142
use[i]
=
-
1
;
143
}
144
}
145
}
146
147
void
go()
148
{
149
//
枚举所有的排列及翻转
150
fmaxArea
=
INF;
151
memset(use,
-
1
,
sizeof
(use));
152
dfs(
0
);
153
out
();
154
}
155
156
int
main()
157
{
158
freopen(
"
packrec.in
"
,
"
r
"
, stdin);
159
freopen(
"
packrec.out
"
,
"
w
"
, stdout);
160
161
for
(
int
i
=
0
; i
<
4
; i
++
)
162
scanf(
"
%d%d
"
,
&
rec[i].w,
&
rec[i].h);
163
go();
164
}
感谢以下链接：
http://sjtulibing.spaces.live.com/blog/cns!2F17193726A8CFC0!139.entry