Matrix
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 271 Accepted Submission(s): 126
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 42
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
题意：对矩阵执行q次 4种类型的操作 输出 最终矩阵
题解：
对于交换行、交换列的操作，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
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<queue>
5 #include<stack>
6 #include<map>
7 #include<set>
8 #include<algorithm>
9 #define LL __int64
10 #define pi acos(-1.0)
11 #define mod 1
12 #define maxn 10000
13 using namespace std;
14 int t;
15 int mp[1005][1005] ;
16 int n,m,q;
17 int a,x,y;
18 int l[1005],h[1005];
19 int ladd[1005],hadd[1005];
20 int main()
21 {
22 scanf("%d",&t);
23 for(int i=1;i<=t;i++)
24 {
25 scanf("%d %d %d",&n,&m,&q);
26 for(int j=1;j<=n;j++)
27 for(int k=1;k<=m;k++)
28 scanf("%d",&mp[j][k]);
29 for(int j=1;j<=n;j++)
30 {
31 h[j]=j;hadd[j]=j;
32 }
33 for(int j=1;j<=m;j++)
34 {
35 l[j]=j; ladd[j]=0;
36 }
37 memset(hadd,0,sizeof(hadd));
38 memset(ladd,0,sizeof(ladd));
39 int t;
40 for(int j=1;j<=q;j++)
41 {
42 scanf("%d %d %d",&a,&x,&y);
43 if(a==1)
44 {
45 t=h[y];
46 h[y]=h[x];
47 h[x]=t;
48 }
49 else
50 if(a==2)
51 {
52 t=l[y];
53 l[y]=l[x];
54 l[x]=t;
55 }
56 else
57 if(a==3)
58 {
59 hadd[h[x]]+=y;
60 }
61 else
62 ladd[l[x]]+=y;
63 }
64 for(int j=1;j<=n;j++)
65 {
66 printf("%d",mp[h[j]][l[1]]+hadd[h[j]]+ladd[l[1]]);
67 for(int k=2;k<=m;k++)
68 {
69 printf(" %d",mp[h[j]][l[k]]+hadd[h[j]]+ladd[l[k]]);
70 }
71 printf("\n");
72 }
73 }
74 return 0;
75 }