Problem Description
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
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
M after all
q
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
行列分开处理即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int T, n, m, q, c, x, y;
int f[1005][1005];
struct change
{
int x, y;
}a[maxn], b[maxn];
int main()
{
scanf("%d", &T);
while (T--)
{
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= n; i++) { a[i].x = i; a[i].y = 0; }
for (int i = 1; i <= m; i++) { b[i].x = i; b[i].y = 0; }
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
scanf("%d", &f[i][j]);
}
}
while (q--)
{
scanf("%d%d%d", &c, &x, &y);
if (c == 1) swap(a[x], a[y]);
if (c == 2) swap(b[x], b[y]);
if (c == 3) a[x].y += y;
if (c == 4) b[x].y += y;
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
printf("%d", f[a[i].x][b[j].x] + a[i].y + b[j].y);
if (j < m) printf(" "); else printf("\n");
}
}
}
return 0;
}
Close