hdu 5671（模拟）

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

23 4 21 2 3 42 3 4 53 4 5 61 1 23 1 102 2 21 1010 11 1 22 1 2

 

Sample Output

12 13 14 151 2 3 43 4 5 61 1010 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 longll 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;}