bzoj2396 神奇的矩阵

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;
}