BZOJ2396 神奇的矩阵 【随机化 + 矩乘】

题目链接

BZOJ2396

题解

一种快速判断两个矩阵是否相等的方法：
对于两个\(n * n\)矩阵，两边同时乘一个\(n * 1\)的随机矩阵，如果结果相等，那么有很大概率两个矩阵相等

如果左边是\(A * B\)的话，用矩阵的结合律先让\(B\)乘就好了，这样子总是一个\(n * n\)的矩阵乘一个\(n * 1\)的矩阵
复杂度\(O(n^2)\)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 1005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn],d[maxn];
int s[maxn],t1[maxn],t2[maxn],n,flag;
void mul(int a[][maxn],int b[maxn],int s[maxn]){
REP(i,n){
s[i] = 0;
REP(j,n) s[i] += a[i][j] * b[j];
}
}
int main(){
while (~scanf("%d",&n)){
flag = true;
REP(i,n) REP(j,n) a[i][j] = read();
REP(i,n) REP(j,n) b[i][j] = read();
REP(i,n) REP(j,n) c[i][j] = read();
REP(i,n) d[i] = rand();
mul(b,d,s);
mul(a,s,t1);
mul(c,d,t2);
REP(i,n) if (t1[i] != t2[i]) {puts("No"); flag = false; break;}
if (flag) puts("Yes");
}
return 0;
}