BZOJ2396 神奇的矩阵

题意：有三个N*N的矩阵a,b,c,判断a*b是否等于c.

思路：暴力判断O(N*3),我没试能不能过。

正解是随机化算法，随机构造列向量p,然后分别计算a*(b*p)和c*p，比较之。

这个过程仅为O(N^2).

随机多组即可。

Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <climits>
#include <cstdlib>
using namespace std;

#define N 1010
struct Matrix {
int w, h, v

;
Matrix() {
memset(v, 0, sizeof v);
}
bool operator == (const Matrix &B) const {
if (w != B.w || h != B.h)
return 0;
register int i, j;
for(i = 0; i < w; ++i)
for(j = 0; j < h; ++j)
if (v[i][j] != B.v[i][j])
return 0;
return 1;
}
void operator = (const Matrix &B) {
w = B.w, h = B.h;
for(int i = 0; i < w; ++i)
for(int j = 0; j < h; ++j)
v[i][j] = B.v[i][j];
}
void operator *= (const Matrix &B);
}A, B, C, ran, tmp1, tmp2, get;

void Matrix::operator *= (const Matrix &B) {
memset(get.v, 0, sizeof get.v);
get.w = w, get.h = B.h;
register int i, j, k;
for(i = 0; i < get.w; ++i)
for(k = 0; k < h; ++k)
for(j = 0; j < get.h; ++j)
get.v[i][j] += v[i][k] * B.v[k][j];
*this = get;
}
int main() {
int n;
register int i, j;
while(scanf("%d", &n) != EOF) {
A.w = A.h = n;
B.w = B.h = n;
C.w = C.h = n;
for(i = 0; i < n; ++i)
for(j = 0; j < n; ++j)
scanf("%d", &A.v[i][j]);
for(i = 0; i < n; ++i)
for(j = 0; j < n; ++j)
scanf("%d", &B.v[i][j]);
for(i = 0; i < n; ++i)
for(j = 0; j < n; ++j)
scanf("%d", &C.v[i][j]);

int wrong = 0;
for(int Case = 1; Case <= 1; ++Case) {
ran.w = 1, ran.h = n;
for(i = 0; i < n; ++i)
ran.v[0][i] = rand();
tmp1 = ran;
tmp2 = ran;
tmp1 *= A, tmp1 *= B;
tmp2 *= C;

if (!(tmp1 == tmp2)) {
wrong = 1;
break;
}
}

if (wrong)
puts("No");
else
puts("Yes");
}

return 0;
}