HDU-5671 Matrix

Matrix
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1378 Accepted Submission(s): 545

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

很有趣的题目考察自己最数组的模拟思想

#include "cstdio"#include "algorithm"using namespace std;const int MAXX=10000+10;int map[MAXX][MAXX];int xa[MAXX];int ya[MAXX];int xc[MAXX];int yc[MAXX];int n,m;int q;int main(){    int T;scanf("%d",&T);    while(T--){        scanf("%d %d %d",&n,&m,&q);        for(int i=1;i<=n;++i)            for(int j=1;j<=m;j++)                scanf("%d",&map[i][j]);        for(int i=1;i<=n;i++) xa[i]=i,xc[i]=0;        for(int i=1;i<=m;i++) ya[i]=i,yc[i]=0;        int a,x,y;        for(int i=1;i<=q;++i){            scanf("%d %d %d",&a,&x,&y);            if(a==1){                swap(xa[x], xa[y]);            }            if(a==2)                swap(ya[x], ya[y]);            if(a==3)                xc[xa[x]]+=y;            if(a==4)                yc[ya[x]]+=y;        }        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                if(j>1)                    printf(" ");                printf("%d",map[xa[i]][ya[j]]+xc[xa[i]]+yc[ya[j]]);            }            printf("\n");        }    }    return 0;}