hdoj--5671--Matrix（思维）



Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1027    Accepted Submission(s): 427

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
题意：n*m的矩阵，进行s次操作，op,x,y,op==1,2表示交换行，列，op==3,4表示行+y,列+y，输出最后产生的矩阵
通过构建新图减少运算的次数
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define MAXN 10100int r[MAXN],c[MAXN],addr[MAXN],addc[MAXN],map[1010][1010];int main(){int t;scanf("%d",&t);while(t--){int n,m,s;scanf("%d%d%d",&n,&m,&s);memset(map,0,sizeof(map));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++){r[i]=i;addr[i]=0;}for(int i=1;i<=m;i++)c[i]=i,addc[i]=0;int op,x,y;for(int i=0;i<s;i++){scanf("%d%d%d",&op,&x,&y);if(op==1)swap(r[x],r[y]);else if(op==2)swap(c[x],c[y]);else if(op==3)addr[r[x]]+=y;else if(op==4)addc[c[x]]+=y;}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(j>1) printf(" ");printf("%d",map[r[i]][c[j]]+addr[r[i]]+addc[c[j]]);}printf("\n");}}return 0;}