Matrix
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 853 Accepted Submission(s): 360
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)
;
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
.
after all q
operations.
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
1 2 3 4
3 4 5 6
1 10
10 1
首先要进行的矩阵变换次数q非常多,
如果每进行一次变换就对矩阵整体进行操作,
势必TLE。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int maxn=1000+10;
int maps[maxn][maxn];
int a[maxn],b[maxn],c[maxn],d[maxn];//a,b数组用来储存当前行或者列是原来第几行或者第几列
//而c,m数组用来储存该行或该列整体加减了多少。
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m,q;
int s,x,y;
scanf("%d%d%d",&n,&m,&q);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
for(int i=1;i<=n;i++)
a[i]=i;
for(int i=1;i<=m;i++)
b[i]=i;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&maps[i][j]);
while(q--)
{
scanf("%d%d%d",&s,&x,&y);
if(s==1)
swap(a[x],a[y]);
if(s==2)
swap(b[x],b[y]);
if(s==3)
c[a[x]]+=y;
if(s==4)
d[b[x]]+=y;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
printf("%d%c",maps[a[i]][b[j]]+c[a[i]]+d[b[j]],(j==m)?'\n':' ');
}
return 0;
}