HDU 5671 Matrix

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 477    Accepted Submission(s): 207


Problem Description

There is a matrix M that
has n rows
and m columns (1≤n≤1000,1≤m≤1000).Then
we perform q(1≤q≤100,000) operations:

1 x y: Swap row x and row y (1≤x,y≤n);

2 x y: Swap column x and column y (1≤x,y≤m);

3 x y: Add y to all elements in row x (1≤x≤n,1≤y≤10,000);

4 x y: Add y to all elements in column x (1≤x≤m,1≤y≤10,000);

 

Input

There are multiple test cases. The first line of input contains an integer T(1≤T≤20) indicating
the number of test cases. For each test case:

The first line contains three integers n, m and q.

The following n lines
describe the matrix M.(1≤Mi,j≤10,000) for
all (1≤i≤n,1≤j≤m).

The following q lines
contains three integers a(1≤a≤4), x and y.

 

Output

For each test case, output the matrix M after
all q operations.

 

Sample Input

2
3 4 2
1 2 3 4
2 3 4 5
3 4 5 6
1 1 2
3 1 10
2 2 2
1 10
10 1
1 1 2
2 1 2

 

Sample Output

12 13 14 15
1 2 3 4
3 4 5 6
1 10
10 1

Hint Recommand to use scanf and printf 题意:给你几种操作，对矩阵进行一系列操作后，输出最后的矩阵。有4种操作:交换x，y列，交换x，y行，x行增加y，x列增加y。思路:对每行每列，记录它实际的行和列，还有增量。最后直接实际的位置加增量就好。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f;
const double  pi = acos(-1.0);
const ll N = 1e3 + 10;
int mat

;
ll ans

;
struct node
{
int root;
ll add;
} line
, cal
;
int main()
{
int t;
cin>>t;
while(t--)
{
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
for(int i = 0; i<n; i++)
line[i].root = i, line[i].add = 0LL;
for(int j = 0; j<m; j++)
cal[j].root = j, cal[j].add =0LL;
for(int i = 0; i<n; i++)
for(int j = 0; j<m; j++)
scanf("%d", &mat[i][j]);
int x, y, a;
while(q--)
{
scanf("%d%d%d", &a, &x, &y);
--x;
if(a == 1)
{
--y;
swap(line[x], line[y]);
}
if(a == 2)
{
--y;
swap(cal[x], cal[y]);
}
if(a == 3)
{
//--y;
line[x].add+=y;
}
if(a == 4)
{
cal[x].add+=y;
}
}
for(int i = 0; i<n; i++)
for(int j = 0; j<m; j++)
{
ans[i][j] = mat[line[i].root][cal[j].root] + line[i].add + cal[j].add;
}
for(int i = 0; i<n; i++)
{
for(int j = 0; j<m; j++)
{
if(j == 0)
printf("%I64d", ans[i][j]);
else printf(" %I64d", ans[i][j]);
}
printf("\n");
}
}
return 0;
}