PM3(矩阵相乘 行和的简便运算)

http://poj.org/problem?id=3213

PM3

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 3146		Accepted: 1108

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 Cwill 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 von the second one, which indicates
the element of C at row r, 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：判断A、B两矩阵相乘的结果C矩阵中是否有唯一错的那个数字，如果没有，输出yes,,否则输出no,并输出错误数字位置和正确答案。

题解：由于有且只有一个位置有错，则可以简化为求出相乘后的矩阵的行和与C比较，若不相同，则找出了有错的行，然后在逐个比较找出错误的位置。技巧就在求行和的时候，矩阵的对应相乘后的行和等于A的某一行的数字分别乘以B的某一行之和，最后求和。 输入矩阵也可以只用一个循环：%s,a[i]；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstdlib>
#include<string>

#define maxn 1010
using namespace std;

int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];
int bcol[maxn],ccol[maxn];
int main()
{
int m,n,p;
int fg=0,f1=0,f2=0,ans;
scanf("%d %d %d",&n,&p,&m);
for(int i=0;i<n;i++)
for(int j=0; j<p; j++)
scanf("%d",&a[i][j]);
for(int i=0;i<p;i++)
for(int j=0; j<m; j++)
scanf("%d",&b[i][j]);
for(int i=0;i<n;i++)
for(int j=0; j<m; j++)
scanf("%d",&c[i][j]);
for(int i=0;i<p;i++)
{
bcol[i]=0;
for(int j=0;j<m;j++)
bcol[i]+=b[i][j];
}
for(int i=0;i<n;i++)
{
ccol[i]=0;
for(int j=0;j<m;j++)
ccol[i]+=c[i][j];
}
for(int i=0;i<n;i++)
{
int temp=0;
for(int j=0;j<p;j++)
temp+=a[i][j]*bcol[j];
if(temp!=ccol[i])
{
fg=1;
f1=i;
break;
}
}
if(!fg)
puts("Yes");
else
{
for(int i=0;i<m;i++)
{
int cnt=0;
for(int j=0;j<p;j++)
{
cnt+=a[f1][j]*b[j][i];
}
if(c[f1][i]!=cnt)
{
f2=i;
ans=cnt;
break;
}
}
puts("No");
printf("%d %d\n",f1+1,f2+1);
printf("%d\n",ans);
}
return 0;
}