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

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 
 
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;}