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bzoj2396 神奇的矩阵

2017-06-14 14:42 253 查看
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2396

【题解】

我们随机一个1*n的矩阵D,根据矩阵乘法的结合律,如果A*B=C,右D*(A*B)=D*C,即(D*A)*B=C,那么矩阵乘法就是O(n^2)的复杂度了。

多随机几次即可。

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

inline int randi(int n) {
return ((rand() << 15) + rand()) % n + 1;
}

int n;

struct mat {
int n, m, a[510][510];
inline void init (int _n, int _m) {
n = _n, m = _m;
memset(a, 0, sizeof a);
}
}A, B, C, o, tA, tB, tC;

mat ta, tb, tc;
inline void mul() {
tc.init(ta.n, tb.m);
for (int i=1; i<=tc.n; ++i)
for (int j=1; j<=tc.m; ++j)
for (int k=1; k<=ta.m; ++k) tc.a[i][j] = tc.a[i][j] + ta.a[i][k] * tb.a[k][j];
}

int main() {
while(~scanf("%d", &n)) {
A.init(n, n); B.init(n, n); C.init(n, n);
for (int i=1; i<=n; ++i)
for (int j=1; j<=n; ++j) scanf("%d", &A.a[i][j]);
for (int i=1; i<=n; ++i)
for (int j=1; j<=n; ++j) scanf("%d", &B.a[i][j]);
for (int i=1; i<=n; ++i)
for (int j=1; j<=n; ++j) scanf("%d", &C.a[i][j]);

for (int T = 1; T <= 10; ++T) {
o.init(1, n);
for (int i=1; i<=n; ++i)
o.a[1][i] = randi(n);

ta = o, tb = C;
mul(); tC = tc;
ta = o, tb = A;
mul(); tA = tc;
ta = tA, tb = B;
mul(); tB = tc;

for (int i=1; i<=n; ++i)
if(tC.a[1][i] != tB.a[1][i]) {
puts("No");
goto END;
}
}
puts("Yes");
END:;
}

return 0;
}


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