hdu 5671 Matrix（矩阵行列式交换）

最新推荐文章于 2021-03-22 11:03:54 发布

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本文介绍了一个涉及矩阵操作的问题，包括行和列的交换、以及对指定行或列元素进行增加的操作。通过使用辅助数组记录每次操作的影响，最终输出经过所有操作后的矩阵状态。

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Matrix

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1410 Accepted Submission(s): 562

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

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

Output

For each test case, output the matrix

M after all

q operations.

Sample Input

Sample Output

12 13 14 15
1 2 3 4
3 4 5 6
1 10
10 1
思路：用a,b,c,d四个数组分别表示交换的行数，列数，行增加的值和列增加的值
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 10010;
int map[N][N];
int a[N] , b[N] , c[N] , d[N];
int main()
{
    int t,n,m,q,k,x,y;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&q);
        for(int i = 1 ; i <= n ; i++)
            a[i] = i;
        for(int j = 1 ; j <= m ; j++)
            b[j] = j;
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
        for(int i = 1 ; i <= n ; i++)
            for(int j = 1 ; j <= m ; j++)
               scanf("%d",&map[i][j]);
        while(q--)
        {
            scanf("%d%d%d",&k,&x,&y);
            if(k == 1) swap(a[x],a[y]);  
            if(k == 2) swap(b[x],b[y]);
            if(k == 3) c[a[x]] += y;
            if(k == 4) d[b[x]] += y;
        }
        for(int i = 1 ; i <= n ; i++){
            for(int j = 1 ; j < m ; j++)
            {
                printf("%d ",map[a[i]][b[j]] + c[a[i]] + d[b[j]]);
            }
            printf("%d\n",map[a[i]][b[m]] + c[a[i]] + d[b[m]]);
        }
    }
    return 0;
}