方阵乘法，矩阵乘法，Strassen 算法——算法作业 2.3，EOJ 1050
方阵相乘
Time Limit:1000MS Memory Limit:30000KB
Description
实现两个n*n 方阵相乘的Strassen 算法，这里假设 n 为 2 的方幂。
Input
第一行为一个正整数N,表示有几组测试数据。
每组测试数据的第一行为一个正整数n(1<=n<=100)，n为2的方幂,表示方阵n*n
接下去的n行表示第一个方阵，每行有n个整数，用空格分开。
再接下去的n行表示第二个方阵，每行有n个整数，用空格分开。
Output
对于每组测试出据，输出n行，每行有n个整数，用空格分开，不能有多余的空格。
Sample Input
1
2
1 2
3 4
5 6
7 8
Sample Output
19 22
43 50
朴素的矩阵乘法
1
#include
<
iostream
>
2#include
<
cstdio
>
3
4
using
namespace
std;
5
6
const
int
L
=
103
;
7
8
int
a[ L ][ L ], b[ L ][ L ], c[ L ][ L ];
9
10
int
main()
{
11
int td, n, i, j, k, tmp;
12 scanf( "%d", &td );
13
while ( td-- ) {
14 scanf( "%d", &n );
15 for ( i = 0; i < n; ++i )
16 for ( j = 0; j < n; ++j )
17 scanf( "%d", &a[ i ][ j ] );
18 for ( i = 0; i < n; ++i )
19 for ( j = 0; j < n; ++j )
20 scanf( "%d", &b[ i ][ j ] );
21 for ( i = 0; i < n; ++i )
22 for ( j = 0; j < n; ++j ) {
23 tmp = 0;
24 for ( k = 0; k < n; ++k )
25 tmp += a[ i ][ k ] * b[ k ][ j ];
26 c[ i ][ j ] = tmp;
27
}
28 for ( i = 0; i < n; ++i ) {
29 printf( "%d", c[ i ][ 0 ] );
30 for ( j = 1; j < n; ++j )
31 printf( " %d", c[ i ][ j ] );
32 printf( "\n" );
33 }
34 }
35 return 0;
36
}
37
#include
<
iostream
>
2
#include
<
cstdio
>
3
4
using
namespace
std;5
6
const
int
L
=
103
;7
8
int
a[ L ][ L ], b[ L ][ L ], c[ L ][ L ];9
10
int
main()
{11
int td, n, i, j, k, tmp;12
scanf( "%d", &td );13
while ( td-- ) {14
scanf( "%d", &n );15
for ( i = 0; i < n; ++i )16
for ( j = 0; j < n; ++j )17
scanf( "%d", &a[ i ][ j ] );18
for ( i = 0; i < n; ++i )19
for ( j = 0; j < n; ++j )20
scanf( "%d", &b[ i ][ j ] );21
for ( i = 0; i < n; ++i )22
for ( j = 0; j < n; ++j ) {23
tmp = 0;24
for ( k = 0; k < n; ++k )25
tmp += a[ i ][ k ] * b[ k ][ j ];26
c[ i ][ j ] = tmp;27
}28
for ( i = 0; i < n; ++i ) {29
printf( "%d", c[ i ][ 0 ] );30
for ( j = 1; j < n; ++j )31
printf( " %d", c[ i ][ j ] );32
printf( "\n" );33
}34
}35
return 0;36
}
37
Strassen 算法
1
#include
<
iostream
>
2#include
<
cstdio
>
3
4
using
namespace
std;
5
6
#define
L 102
7
#define
LIM 400
8
9typedef
int
Mat[ L ][ L ];
10
11Mat buf[ LIM ];
12
int
top;
13
14
void
input(
int
a[][L],
int
n )
{
15 int i, j;
16 for ( i = 1; i <= n; ++i ) {
17 for ( j = 1; j <= n; ++j ) {
18 scanf( "%d", &a[ i ][ j ] );
19 }
20 }
21}
22
23
void
output(
int
c[][L],
int
n )
{
24 int i, j;
25 for ( i = 1; i <= n; ++i ) {
26 for ( j = 1; j < n; ++j ) {
27 printf( "%d ", c[ i ][ j ] );
28 }
29 printf( "%d\n", c[ i ][ j ] );
30 }
31}
32
33
void
get
(
int
a[][L],
int
a11[][L],
int
a12[][L],
int
a21[][L],
int
a22[][L],
int
n )
{
34 int i, j;
35 for ( i = 1; i <= n; ++i ) {
36 for ( j = 1; j <= n; ++j ) {
37 a11[ i ][ j ] = a[ i ][ j ];
38 a12[ i ][ j ] = a[ i ][ j + n ];
39 a21[ i ][ j ] = a[ i + n ][ j ];
40 a22[ i ][ j ] = a[ i + n ][ j + n ];
41 }
42 }
43}
44
45
void
put(
int
a[][L],
int
a11[][L],
int
a12[][L],
int
a21[][L],
int
a22[][L],
int
n )
{
46 int i, j;
47 for ( i = 1; i <= n; ++i ) {
48 for ( j = 1; j <= n; ++j ) {
49 a[ i ][ j ] = a11[ i ][ j ];
50 a[ i ][ j + n ] = a12[ i ][ j ];
51 a[ i + n ][ j ] = a21[ i ][ j ];
52 a[ i + n ][ j + n ] = a22[ i ][ j ];
53 }
54 }
55}
56
57
void
add(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{
58 int i, j;
59 for ( i = 1; i <= n; ++i ) {
60 for ( j = 1; j <= n; ++j ) {
61 c[ i ][ j ] = a[ i ][ j ] + b[ i ][ j ];
62 }
63 }
64}
65
66
void
sub(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{
67 int i, j;
68 for ( i = 1; i <= n; ++i ) {
69 for ( j = 1; j <= n; ++j ) {
70 c[ i ][ j ] = a[ i ][ j ] - b[ i ][ j ];
71 }
72 }
73}
74
75
void
mul(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{
76#define ADD(m) Mat &m = buf[ top++ ]
77#define ADDS(a) ADD(a##11); ADD(a##12); ADD(a##21); ADD(a##22)
78#define ENTER ADDS(a); ADDS(b); ADDS(c); ADD(d1); ADD(d2); ADD(d3); ADD(d4); ADD(d5); ADD(d6); ADD(d7); ADD(t1); ADD(t2)
79#define LEAVE top -= 21
80
81 ENTER;
82
83 if ( top >= LIM ) {
84 // for debug
85 fprintf( stderr, "buf overflow!!" );
86 LEAVE;
87 return;
88 }
89
90
91 if ( n < 1 ) {
92 LEAVE;
93 return;
94 }
95 if ( n == 1 ) {
96 c[ 1 ][ 1 ] = a[ 1 ][ 1 ] * b[ 1 ][ 1 ];
97 LEAVE;
98 return;
99 }
100 n >>= 1;
101 get( a, a11, a12, a21, a22, n );
102 get( b, b11, b12, b21, b22, n );
103
104 add( t1, a11, a22, n );
105 add( t2, b11, b22, n );
106 mul( d1, t1, t2, n );
107
108 add( t1, a21, a22, n );
109 mul( d2, t1, b11, n );
110
111 sub( t2, b12, b22, n );
112 mul( d3, a11, t2, n );
113
114 sub( t2, b21, b11, n );
115 mul( d4, a22, t2, n );
116
117 add( t1, a11, a12, n );
118 mul( d5, t1, b22, n );
119
120 sub( t1, a21, a11, n );
121 add( t2, b11, b12, n );
122 mul( d6, t1, t2, n );
123
124 sub( t1, a12, a22, n );
125 add( t2, b21, b22, n );
126 mul( d7, t1, t2, n );
127
128 add( t1, d1, d4, n );
129 sub( t2, d5, d7, n );
130 sub( c11, t1, t2, n );
131
132 add( c12, d3, d5, n );
133
134 add( c21, d2, d4, n );
135
136 add( t1, d1, d3, n );
137 sub( t2, d2, d6, n );
138 sub( c22, t1, t2, n );
139
140 put( c, c11, c12, c21, c22, n );
141
142 LEAVE;
143}
144
145
int
main()
{
146 int td, n, a[ L ][ L ], b[ L ][ L ], c[ L ][ L ];
147 scanf( "%d", &td );
148 while ( td-- > 0 ) {
149 top = 0;
150 scanf( "%d", &n );
151 input( a, n );
152 input( b, n );
153 mul( c, a, b, n );
154 output( c, n );
155 }
156 return 0;
157}
158
#include
<
iostream
>
2
#include
<
cstdio
>
3
4
using
namespace
std;5
6
#define
L 102
7
#define
LIM 400
8
9
typedef
int
Mat[ L ][ L ];10
11
Mat buf[ LIM ];12
int
top;13
14
void
input(
int
a[][L],
int
n )
{15
int i, j;16
for ( i = 1; i <= n; ++i ) {17
for ( j = 1; j <= n; ++j ) {18
scanf( "%d", &a[ i ][ j ] );19
}20
}21
}
22
23
void
output(
int
c[][L],
int
n )
{24
int i, j;25
for ( i = 1; i <= n; ++i ) {26
for ( j = 1; j < n; ++j ) {27
printf( "%d ", c[ i ][ j ] );28
}29
printf( "%d\n", c[ i ][ j ] );30
}31
}
32
33
void
get
(
int
a[][L],
int
a11[][L],
int
a12[][L],
int
a21[][L],
int
a22[][L],
int
n )
{34
int i, j;35
for ( i = 1; i <= n; ++i ) {36
for ( j = 1; j <= n; ++j ) {37
a11[ i ][ j ] = a[ i ][ j ];38
a12[ i ][ j ] = a[ i ][ j + n ];39
a21[ i ][ j ] = a[ i + n ][ j ];40
a22[ i ][ j ] = a[ i + n ][ j + n ];41
}42
}43
}
44
45
void
put(
int
a[][L],
int
a11[][L],
int
a12[][L],
int
a21[][L],
int
a22[][L],
int
n )
{46
int i, j;47
for ( i = 1; i <= n; ++i ) {48
for ( j = 1; j <= n; ++j ) {49
a[ i ][ j ] = a11[ i ][ j ];50
a[ i ][ j + n ] = a12[ i ][ j ];51
a[ i + n ][ j ] = a21[ i ][ j ];52
a[ i + n ][ j + n ] = a22[ i ][ j ];53
}54
}55
}
56
57
void
add(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{58
int i, j;59
for ( i = 1; i <= n; ++i ) {60
for ( j = 1; j <= n; ++j ) {61
c[ i ][ j ] = a[ i ][ j ] + b[ i ][ j ];62
}63
}64
}
65
66
void
sub(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{67
int i, j;68
for ( i = 1; i <= n; ++i ) {69
for ( j = 1; j <= n; ++j ) {70
c[ i ][ j ] = a[ i ][ j ] - b[ i ][ j ];71
}72
}73
}
74
75
void
mul(
int
c[][L],
int
a[][L],
int
b[][L],
int
n )
{76
#define ADD(m) Mat &m = buf[ top++ ]77
#define ADDS(a) ADD(a##11); ADD(a##12); ADD(a##21); ADD(a##22)78
#define ENTER ADDS(a); ADDS(b); ADDS(c); ADD(d1); ADD(d2); ADD(d3); ADD(d4); ADD(d5); ADD(d6); ADD(d7); ADD(t1); ADD(t2)79
#define LEAVE top -= 2180
81
ENTER;82
83
if ( top >= LIM ) {84
// for debug85
fprintf( stderr, "buf overflow!!" );86
LEAVE;87
return;88
}89
90
91
if ( n < 1 ) {92
LEAVE;93
return;94
}95
if ( n == 1 ) {96
c[ 1 ][ 1 ] = a[ 1 ][ 1 ] * b[ 1 ][ 1 ];97
LEAVE;98
return;99
}100
n >>= 1;101
get( a, a11, a12, a21, a22, n );102
get( b, b11, b12, b21, b22, n );103
104
add( t1, a11, a22, n );105
add( t2, b11, b22, n );106
mul( d1, t1, t2, n );107
108
add( t1, a21, a22, n );109
mul( d2, t1, b11, n );110
111
sub( t2, b12, b22, n );112
mul( d3, a11, t2, n );113
114
sub( t2, b21, b11, n );115
mul( d4, a22, t2, n );116
117
add( t1, a11, a12, n );118
mul( d5, t1, b22, n );119
120
sub( t1, a21, a11, n );121
add( t2, b11, b12, n );122
mul( d6, t1, t2, n );123
124
sub( t1, a12, a22, n );125
add( t2, b21, b22, n );126
mul( d7, t1, t2, n );127
128
add( t1, d1, d4, n );129
sub( t2, d5, d7, n );130
sub( c11, t1, t2, n );131
132
add( c12, d3, d5, n );133
134
add( c21, d2, d4, n );135
136
add( t1, d1, d3, n );137
sub( t2, d2, d6, n );138
sub( c22, t1, t2, n );139
140
put( c, c11, c12, c21, c22, n );141
142
LEAVE;143
}
144
145
int
main()
{146
int td, n, a[ L ][ L ], b[ L ][ L ], c[ L ][ L ];147
scanf( "%d", &td );148
while ( td-- > 0 ) {149
top = 0;150
scanf( "%d", &n );151
input( a, n );152
input( b, n );153
mul( c, a, b, n );154
output( c, n );155
}156
return 0;157
}
158
我的实现有点丑。。。