Time Limit: 3000/1500 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 891Accepted Submission(s): 371
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
Source
BestCoder Round #81 (div.2)
题目大意：
有一个nn行mm列的矩阵(1 \leq n \leq 1000 ,1 \leq m \leq 1000 )(1≤n≤1000,1≤m≤1000)，在这个矩阵上进行qq (1 \leq q \leq 100,000)(1≤q≤100,000) 个操作：
1 x y: 交换矩阵MM的第xx行和第yy行(1 \leq x,y \leq n)(1≤x,y≤n);
2 x y: 交换矩阵MM的第xx列和第yy列(1 \leq x,y \leq m)(1≤x,y≤m);
3 x y: 对矩阵MM的第xx行的每一个数加上y(1 \leq x \leq n,1 \leq y \leq 10,000)y(1≤x≤n,1≤y≤10,000);
4 x y: 对矩阵MM的第xx列的每一个数加上y(1 \leq x \leq m,1 \leq y \leq 10,000)y(1≤x≤m,1≤y≤10,000);对于每组数据，输出经过所有q个操作之后的矩阵M。
解题思路：如果单纯进行模拟的话，数据量比较大，所以采用标记的方式。分别记录当前状态下每一行、每一列是原始数组的哪一行、哪一列即可。对每一行、每一列加一个数的操作，也可以两个数组分别记录。注意当交换行、列的同时，也要交换增量数组。输出时通过索引找到原矩阵中的值，再加上行、列的增量。
详见代码。
#include
#include
#include
using namespace std; int Map[1010][1010],a; int h[1010],l[1010],dh[1010],dl[1010]; int main() { int T; scanf("%d",&T); while (T--) { memset(dh,0,sizeof(dh)); memset(dl,0,sizeof(dl)); int n,m,q; scanf("%d%d%d",&n,&m,&q); for (int i=1; i<=n; i++) { for (int j=1; j<=m; j++) { scanf("%d",&Map[i][j]); h[i]=i; l[j]=j; } } for (int i=0; i