Matrix
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
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
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
1 2 3 4
3 4 5 6
1 10
10 1
Hint
Recommand to use scanf and printf
Source
思路：就是标记记录那行是原数组的那行，在原数组的那行加减就好；
代码
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
ll a[][];
ll hang[],lie[],hangadd[],lieadd[];//不管你怎么换 ，都不可能把这行给拆了
int main()
{
ll casee,i,t;
scanf("%I64d",&casee);
while(casee--)
{
ll n,m,q;
memset(hangadd,,sizeof(hangadd));
memset(lieadd,,sizeof(lieadd));
scanf("%I64d%I64d%I64d",&n,&m,&q);
for(i=;i<=n;i++)
for(t=;t<=m;t++)
{
scanf("%I64d",&a[i][t]);
hang[i]=i;
lie[t]=t;
}
while(q--)
{
ll op,han,li;
scanf("%I64d%I64d%I64d",&op,&han,&li);
if(op==)
{
swap(hang[han],hang[li]);
}
else if(op==)
{
swap(lie[han],lie[li]);
}
else if(op==)
hangadd[hang[han]]+=li;
else
lieadd[lie[han]]+=li;
}
for(i=;i<=n;i++)
for(t=;t<=m;t++)
printf("%I64d%c",a[hang[i]][lie[t]]+hangadd[hang[i]]+lieadd[lie[t]],t==m?'\n':' ');
}
return ;
}