POJ 3213 PM3 矩阵乘法优化

PM³

Time Limit: 5000MS Memory Limit: 131072KTotal 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 21 2 -13 -1 0-1 00 21 3-2 -1-3 -2

Sample Output

No。1 21

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[N][N],B[N][N],C[N][N];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[N][N],B[N][N],C[N][N];int b_c[N],c_c[N];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;}