hust校赛 1615 Matrix 矩阵和其逆矩阵的乘积中不为0的元素个数
Matrix
Time Limit: 3 Sec Memory Limit: 128 MBSubmissions: 164 Solved: 38
Description
To efficient calculate the multiplication of a sparse matrix is very useful in industrial filed.
Let's consider this problem:
A is an N*N matrix which only contains 0 or 1. And we want to know the result of A*AT.
Formally, we define B = A*AT, A(i,j) is equal to 1 or 0, and we know the number of 1 in matrix A is M
And your task is to calculate B.
Input
The input contains several test cases. The first line of input contains a integer C indicating the number of the cases.
For each test case, the first line contains two integer N and M.
and each of next M lines contains two integer X and Y, which means A(x,y) is 1.
N <= 100,000 M <= 1000.C <= 10
Output
For each test case, it should have a integer W indicating how many element in Matrix B isn't zero in one line.
Sample Input
2 5 3 1 0 2 1 3 3 3 3 0 0 1 0 2 0
Sample Output
3 9
HINT
AT means the Transpose of matrix A, for more details, AT(i,j) = A(j,i).
eg:
if Matrix A is:
1 2 3
4 5 6
7 8 9
then the matrix AT is
1 4 7
2 5 8
3 6 9
Source
The 7th(2012) ACM Programming Contest of HUST
Problem Setter: Zheng Zhang
题意:矩阵A为0,1矩阵，求矩阵B等于A乘以A的逆，然后求B有多少个不等于0的数
思路：
在本子上写几组数据即可发现下面的规律 ：
#include
#include
#include
using namespace std;
int x[1000+10],y[1000+10];
int main()
{
int cas,i,j,n,m,xx,yy,cnt;
scanf("%d",&cas);
while(cas--)
{
scanf("%d %d",&n,&m);
for(i=0;i 同学的解题报告：
Problem I: Matrix
http://acm.hust.edu.cn/problem.php?cid=1106&pid=8
/*
题意:矩阵A为0,1矩阵，求矩阵B等于A乘以A的逆，然后求B有多少个1
思路：矩阵乘以矩阵的逆相当于A每行和所有的行相乘。所以，比较矩阵A每列对应有多少个相等的1，
循环A的行，看第i行的每列最大有多少个数"1",就是对应B的第i行的1的个数。但是如果用二维数组
开10000*10000会超内存，而且是1的也最多只有1000个，所以用op数组标记，用row[i][]对应的所有
是1的列存起来，col[]为第j列有多少 个1.然后查找就可以了
*/
#include
#include
int row[1005][1005];
int op[100005];
int col[100005];
int Max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,m,T,i,x,y,t,sum,w,v;
scanf("%d",&T);
while(T--)
{
memset(row,0,sizeof(row));
memset(col,0,sizeof(col));
memset(op,0,sizeof(op));
scanf("%d%d",&n,&m);
v=1;
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
if(op[x])
{
w=op[x];
t=++row[w][0];
row[w][t]=y;
col[y]++;
}
else
{
op[x]=v;
row[v][1]=y;
row[v][0]=1;
col[y]++;
v++;
}
}
sum=0;
for(i=1;i