Matrix

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 768 Accepted Submission(s): 323

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)

//开四个数组记录行列变换。。还有相加的值

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll ma[1005][1005];
ll row[1005],col[1005],sumr[1005],sumc[1005];
int main()
{int t;scanf("%d",&t);while(t--){ll n,m,q;scanf("%lld%lld%lld",&n,&m,&q);memset(sumr,0,sizeof(sumr));memset(sumc,0,sizeof(sumc));for(int i=1;i<=n;i++){row[i]=i;for(int k=1;k<=m;k++){scanf("%lld",&ma[i][k]);col[k]=k;}}while(q--){ll type,x,y;scanf("%lld%lld%lld",&type,&x,&y);if(type==1){swap(row[x],row[y]);swap(sumr[x],sumr[y]);}else if(type==2){swap(col[x],col[y]);swap(sumc[x],sumc[y]);}else if(type==3){sumr[x]+=y;}else{sumc[x]+=y;}}for(int i=1;i<=n;i++){for(int k=1;k<=m-1;k++){ll t=ma[row[i]][col[k]]+sumr[i]+sumc[k];printf("%lld ",t);}printf("%lld\n",ma[row[i]][col[m]]+sumr[i]+sumc[m]);}}return 0;
}

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