USACOTrainning.Mother's Milk
这道题蛮有意思的，大意是有三个水桶ABC，只有C水桶放满水，从一个水桶倒到另一个水桶，要么一个桶空，要么一个桶满为止，问说当A桶空时，C桶可能出现的所有的可能状态。
比较容易想到的是讲(a,b,c)表示为状态，然后暴力搜索出所有的可能性，因为一共就3个桶，操作也不复杂，对使用过的状态进行标记，这样可以在O(N3)完成，因为USACO的数据很小,<=20,所以就直接AC掉了，看了ANALYSIS之后知道，其实状态用N2就可以搞定，这样O(N2)就可以完成了。
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
#include
<
queue
>
9
using
namespace
std;
10
11
const
int
MAX
=
1005
;
12
13
struct
WATER
14
{
15
int
a, c;
16
WATER(
int
aa,
int
cc) : a(aa), c(cc) {}
17
WATER(){}
18
};
19
20
int
A, B, C;
21
queue
<
WATER
>
water;
22
vector
<
int
>
ans;
23
bool
use[MAX][MAX];
24
//
存a和c的水量就行b = C - a - c
25
//
状态只有O(n*n)
26
27
void
pour(
int
&
x,
int
&
y,
int
X,
int
Y)
28
{
29
if
(x
<=
Y
-
y)
30
{
31
y
+=
x;
32
x
=
0
;
33
}
34
else
35
{
36
x
-=
(Y
-
y);
37
y
=
Y;
38
}
39
}
40
41
void
judge(
int
a,
int
b,
int
c)
42
{
43
if
(use[a][c]
==
false
)
44
{
45
use[a][c]
=
true
;
46
water.push(WATER(a, c));
47
}
48
}
49
50
void
check(
int
a,
int
b,
int
c)
51
{
52
int
aa
=
a;
53
int
bb
=
b;
54
int
cc
=
c;
55
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pour(a, b, A, B);
57
judge(a, b, c);
58
a
=
aa, b
=
bb, c
=
cc;
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60
pour(a, c, A, C);
61
judge(a, b, c);
62
a
=
aa, b
=
bb, c
=
cc;
63
64
pour(b, a, B, A);
65
judge(a, b, c);
66
a
=
aa, b
=
bb, c
=
cc;
67
68
pour(b, c, B, C);
69
judge(a, b, c);
70
a
=
aa, b
=
bb, c
=
cc;
71
72
pour(c, a, C, A);
73
judge(a, b, c);
74
a
=
aa, b
=
bb, c
=
cc;
75
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pour(c, b, C, B);
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judge(a, b, c);
78
a
=
aa, b
=
bb, c
=
cc;
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}
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void
go()
82
{
83
while
(
!
water.empty())
84
{
85
WATER w
=
water.front();
86
if
(w.a
==
0
) ans.push_back(w.c);
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water.pop();
88
check(w.a, C
-
w.a
-
w.c, w.c);
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}
90
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sort(ans.begin(), ans.end());
92
for
(
int
i
=
0
; i
<
ans.size(); i
++
)
93
{
94
if
(i) printf(
"
%d
"
, ans[i]);
95
else
printf(
"
%d
"
, ans[i]);
96
}
97
printf(
"
\n
"
);
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}
99
100
int
main()
101
{
102
freopen(
"
milk3.in
"
,
"
r
"
, stdin);
103
freopen(
"
milk3.out
"
,
"
w
"
, stdout);
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105
int
a, b, c;
106
scanf(
"
%d%d%d
"
,
&
A,
&
B,
&
C);
107
water.push(WATER(
0
, C));
108
memset(use,
0
,
sizeof
(use));
109
use[
0
][C]
=
true
;
110
go();
111
}