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UVA - 10895| Time Limit:3000MS | Memory Limit:Unknown | 64bit IO Format:%lld & %llu |
Description
| A: Matrix Transpose |

The individual elements of a matrix are usually given lowercase symbols and are distinguished by subscripts. The
th row and
th column of matrix
is usually referred to as
. For example,
. Matrix subscripts are 1-based.
The transpose of a matrix
, denoted
, is formed by interchanging the rows and columns of
. That is, the
th element of
is the
th element of
. For example, the transpose of matrix
above is:

A matrix is said to be sparse if there are relatively few non-zero elements. As a
-by-
matrix has
number of elements, storing all elements of a large sparse matrix may be inefficient as there
would be many zeroes. There are a number of ways to represent sparse matrices, but essentially they are all the same: store only the non-zero elements of the matrix along with their row and column.
You are to write a program to output the transpose of a sparse matrix of integers.
4 3 3 1 2 3 1 3 2 2 2 3 4 -1 0 3 1 2 3 5 -2 11Note that for a row with all zero elements, the corresponding set would just be one number, `0‘, in the first line, followed by a blank line.
You may assume:
4 3 3 1 2 3 1 3 2 2 2 3 4 -1 0 3 1 2 3 5 -2 11
3 4 2 1 4 1 5 3 1 2 4 3 4 -2 3 1 2 4 2 -1 11
题意:首先给你n*m的矩阵,然后给出每行的情况。第一个数r代表该行有几个非0的数,位置是哪里,然后给出每个位置的值,求矩阵的倒置
思路:用两个vector,一个记录该列有效值所对应的行,还一个记录该位置的值
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
const int MAXN = 10010;
vector<int> row[MAXN];
vector<int> val[MAXN];
int n, m, arr[MAXN];
void print() {
printf("%d %d\n", m, n);
for (int i = 1; i <= m; i++) {
int len = row[i].size();
printf("%d", len);
for (int j = 0; j < len; j++)
printf(" %d", row[i][j]);
if (len == 0)
printf("\n\n");
else {
printf("\n%d", val[i][0]);
for (int j = 1; j < len; j++)
printf(" %d", val[i][j]);
printf("\n");
}
}
}
int main() {
while (scanf("%d%d", &n ,&m) != EOF) {
for (int i = 0; i < MAXN; i++) {
row[i].clear();
val[i].clear();
}
int r, x;
for (int i = 1; i <= n; i++) {
scanf("%d", &r);
for (int j = 1; j <= r; j++)
scanf("%d", &arr[j]);
for (int j = 1; j <= r; j++) {
scanf("%d", &x);
row[arr[j]].push_back(i);
val[arr[j]].push_back(x);
}
}
print();
}
return 0;
}UVA - 10895 Matrix Transpose,布布扣,bubuko.com
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原文地址:http://blog.csdn.net/u011345136/article/details/37885419