POJ 3213 PM3 矩阵乘法优化

PM3

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 3298		Accepted: 1151

Description

USTC has recently developed the Parallel Matrix Multiplication Machine – PM3, which is used for very large matrix multiplication.

Given two matrices A and B, where A is an N × P matrix and B is a P × M matrix, PM3 can compute matrix C = AB in O(P(N + P + M))
time. However the developers of PM3 soon discovered a small problem: there is a small chance that PM3 makes a mistake, and whenever a mistake occurs, the resultant matrix C will contain exactly one incorrect element.

The developers come up with a natural remedy. After PM3 gives the matrix C, they check and correct it. They think it is a simple task, because there will be at most one incorrect element.

So you are to write a program to check and correct the result computed by PM3.

Input

The first line of the input three integers N, P and M (0 < N, P, M ≤ 1,000), which indicate the dimensions of A and B. Then follow N lines with P integers each, giving
the elements of A in row-major order. After that the elements of B and C are given in the same manner.

Elements of A and B are bounded by 1,000 in absolute values which those of C are bounded by 2,000,000,000.

Output

If C contains no incorrect element, print “

Yes

”. Otherwise print “

No

” followed by two more lines, with two integers r and c on the first one, and another integer v on the second one, which indicates
the element of C at rowr, column c should be corrected to v.

Sample Input

2 3 2
1 2 -1
3 -1 0
-1 0
0 2
1 3
-2 -1
-3 -2

Sample Output

No。
1 2
1

Hint

The test set contains large-size input. Iostream objects in C++ or Scanner in Java might lead to efficiency problems.

题意：给矩阵A,B,C。C == A*B，但是C中可能有一个数据是错的。没有错误输出Yes，有就输出No并输出几行几列和正确答案。

解法，暴力矩阵乘法可过，直接写就行

另一种，写矩阵乘法的时候，会发现A的每个数都会乘B中某一行的每个数字。这样，可以处理B的每一行的和，求的A*B的每一行的和与C的每一行的和进行比较，如果相等就没有错误数据，不想等就有错误数据，对出现错误的这一行与B进行矩阵乘法找出那个数据就行了。感觉还是很巧妙的。

暴力的写法已注释

/*暴力求解
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>

using namespace std;

const int N = 1000+10;

int A

,B

,C

;

int main()
{
int n,p,m;
scanf("%d%d%d",&n,&p,&m);
for(int i = 1;i <= n;i++) for(int j = 1;j <= p;j++) scanf("%d",&A[i][j]);
for(int i = 1;i <= p;i++) for(int j = 1;j <= m;j++) scanf("%d",&B[i][j]);
for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) scanf("%d",&C[i][j]);
int tmp;
for(int i = 1;i <= n;i++){
for(int j = 1;j <= m;j++){
tmp = 0;
for(int k = 1;k <= p;k++)
tmp += A[i][k]*B[k][j];
if(tmp != C[i][j]){
printf("No\n%d %d\n%d\n",i,j,tmp);
return 0;
}
}
}
printf("Yes\n");
return 0;
}*/

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>

using namespace std;
const int N = 1000+10;
int A

,B

,C

;
int b_c
,c_c
;
int main(void)
{
int n,p,m;
memset(b_c,0,sizeof b_c);
memset(c_c,0,sizeof c_c);
scanf("%d%d%d",&n,&p,&m);
for(int i = 1;i <= n;i++) for(int j = 1;j <= p;j++) scanf("%d",&A[i][j]);
for(int i = 1;i <= p;i++) for(int j = 1;j <= m;j++) scanf("%d",&B[i][j]),b_c[i] += B[i][j];
for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) scanf("%d",&C[i][j]),c_c[i] += C[i][j];
bool flag = false;
int col; ///记录哪一行出错
for(int i = 1;i <= n;i++){     ///查找哪一行出错
int tmp = 0;
for(int j = 1;j <= p;j++){
tmp += A[i][j]*b_c[j];
}
if(tmp != c_c[i]) {col = i;flag = true;break;}
}
if(!flag) puts("Yes");
else{
puts("No");
for(int i = 1;i <= m;i++){ ///进行矩阵乘法查找就行
int tmp = 0;
for(int j = 1;j <= p;j++){
tmp += A[col][j]*B[j][i];
}
if(tmp != C[col][i]){
printf("%d %d\n%d\n",col,i,tmp);
//return 0;
}
}
}
return 0;
}