HDU 5671 Matrix

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 477    Accepted Submission(s): 207

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 
 
题意:给你几种操作，对矩阵进行一系列操作后，输出最后的矩阵。有4种操作:交换x，y列，交换x，y行，x行增加y，x列增加y。
思路:对每行每列，记录它实际的行和列，还有增量。最后直接实际的位置加增量就好。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f;
const double  pi = acos(-1.0);
const ll N = 1e3 + 10;
int mat[N][N];
ll ans[N][N];
struct node
{
    int root;
    ll add;
} line[N], cal[N];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n, m, q;
        scanf("%d%d%d", &n, &m, &q);
        for(int i = 0; i<n; i++)
            line[i].root = i, line[i].add = 0LL;
        for(int j = 0; j<m; j++)
            cal[j].root = j, cal[j].add =0LL;
        for(int i = 0; i<n; i++)
            for(int j = 0; j<m; j++)
                scanf("%d", &mat[i][j]);
        int x, y, a;
        while(q--)
        {
            scanf("%d%d%d", &a, &x, &y);
            --x;
            if(a == 1)
            {
                --y;
                swap(line[x], line[y]);
            }
            if(a == 2)
            {
                --y;
                swap(cal[x], cal[y]);
            }
            if(a == 3)
            {
                //--y;
                line[x].add+=y;
            }
            if(a == 4)
            {
                cal[x].add+=y;
            }
        }
        for(int i = 0; i<n; i++)
            for(int j = 0; j<m; j++)
        {
            ans[i][j] = mat[line[i].root][cal[j].root] + line[i].add + cal[j].add;
        }
        for(int i = 0; i<n; i++)
        {
            for(int j = 0; j<m; j++)
            {
                if(j == 0)
                    printf("%I64d", ans[i][j]);
                else printf(" %I64d", ans[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}