USACO 1.4 Mother's Milk
比较经典的3个杯子互相倒水的问题。
状态即3个杯子中的水量，用一个3维数组记录，转移为六种组合。
code
1
/*
2
ID: superbi1
3
LANG: C
4
TASK: milk3
5
*/
6
//
BFS
7
//
三个杯子A,B,C 刚开始，C中milk满的，然后根据规则倒，问A为空时的C的milk量的序列
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#include
<
stdio.h
>
9
#include
<
string
.h
>
10
#include
<
stdlib.h
>
11
12
//
其实可以用三维数组记录状态
13
char
flg[
1000000
];
14
15
main()
16
{
17
FILE
*
fin
=
fopen(
"
milk3.in
"
,
"
r
"
);
18
FILE
*
fout
=
fopen(
"
milk3.out
"
,
"
w
"
);
19
int
a, b, c;
20
int
milk[
21
];
21
int
start
=
c;
22
int
que[
1000
], top
=
0
, mk;
23
int
cur;
24
int
I, a0, b0, c0, a1, b1, c1;
25
fscanf(fin,
"
%d%d%d
"
,
&
a,
&
b,
&
c);
26
memset(flg,
0
,
sizeof
(flg));
27
que[top
++
]
=
c;
28
flg[c]
=
1
;
29
for
(I
=
0
; I
<
21
; I
++
) milk[I]
=
0
;
30
while
(top
>
0
) {
31
cur
=
que[
--
top];
32
if
(
!
(cur
/
10000
)) {
33
milk[cur
%
100
]
=
1
;
34
}
35
a0
=
cur
/
10000
;
36
b0
=
(cur
%
10000
)
/
100
;
37
c0
=
cur
%
100
;
38
a1
=
a0;
39
b1
=
b0;
40
c1
=
c0;
41
//
枚举六种倒的情况
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if
(a1
>
0
) {
43
if
(b1
<
b) {
44
if
(b
-
b1
<
a1) {
45
a1
-=
b
-
b1;
46
b1
=
b;
47
}
else
{
48
b1
+=
a1;
49
a1
=
0
;
50
}
51
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
52
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
53
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
54
}
55
}
56
a1
=
a0;
57
b1
=
b0;
58
c1
=
c0;
59
if
(c1
<
c) {
60
if
(c
-
c1
<
a1) {
61
a1
-=
c
-
c1;
62
c1
=
c;
63
}
else
{
64
c1
+=
a1;
65
a1
=
0
;
66
}
67
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
68
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
69
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
70
}
71
}
72
}
73
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a1
=
a0;
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b1
=
b0;
76
c1
=
c0;
77
if
(b1
>
0
) {
78
if
(a1
<
a) {
79
if
(a
-
a1
<
b1) {
80
b1
-=
a
-
a1;
81
a1
=
a;
82
}
else
{
83
a1
+=
b1;
84
b1
=
0
;
85
}
86
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
87
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
88
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
89
}
90
}
91
a1
=
a0;
92
b1
=
b0;
93
c1
=
c0;
94
if
(c1
<
c) {
95
if
(c
-
c1
<
b1) {
96
b1
-=
c
-
c1;
97
c1
=
c;
98
}
else
{
99
c1
+=
b1;
100
b1
=
0
;
101
}
102
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
103
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
104
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
105
}
106
}
107
}
108
109
a1
=
a0;
110
b1
=
b0;
111
c1
=
c0;
112
if
(c1
>
0
) {
113
if
(a1
<
a) {
114
if
(a
-
a1
<
c1) {
115
c1
-=
a
-
a1;
116
a1
=
a;
117
}
else
{
118
a1
+=
c1;
119
c1
=
0
;
120
}
121
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
122
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
123
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
124
}
125
}
126
a1
=
a0;
127
b1
=
b0;
128
c1
=
c0;
129
if
(b1
<
b) {
130
if
(b
-
b1
<
c1) {
131
c1
-=
b
-
b1;
132
b1
=
b;
133
}
else
{
134
b1
+=
c1;
135
c1
=
0
;
136
}
137
if
(
!
flg[a1
*
10000
+
b1
*
100
+
c1]) {
138
flg[a1
*
10000
+
b1
*
100
+
c1]
=
1
;
139
que[top
++
]
=
a1
*
10000
+
b1
*
100
+
c1;
140
}
141
}
142
}
143
}
144
mk
=
0
;
145
for
(I
=
0
; I
<
21
; I
++
) {
146
if
(milk[I]) {
147
if
(
!
mk) { fprintf(fout,
"
%d
"
, I); mk
=
1
; }
148
else
fprintf(fout,
"
%d
"
, I);
149
}
150
}
151
fprintf(fout,
"
\n
"
);
152
exit(
0
);
153
}