題意:有三個N*N的矩陣a,b,c,判斷a*b是否等於c.
思路:暴力判斷O(N*3),我沒試能不能過。
正解是隨機化算法,隨機構造列向量p,然後分別計算a*(b*p)和c*p,比較之。
這個過程僅爲O(N^2).
隨機多組即可。
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[N][N];
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;
}