【hdoj5671】Matrix

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1411    Accepted Submission(s): 563

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 
   
    
  

 

Source

BestCoder Round #81 (div.2)

 

Recommend

wange2014

 

这道题的思想非常好，这就解决了超时的问题。

首相想到的是直接暴力操作，可是这样的复杂度肯定是要超时的。所以我们不对每行每列一一操作，而是对整行整列操作，开两个数组分别记录行下标及列下表，再开两个数组记录每行加多少及每列加多少，输出的时候输出处理后的下标最终的值。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define LCA(s,x) memset(s,0,sizeof(s))
using namespace std;
const int N = 1005;
int map[N][N],a[N],b[N],c[N],d[N];
int main() {
	int T;
	scanf("%d",&T);
	while(T--) {
		int n,m,q;
		LCA(b,0);
		LCA(d,0);
		for(int i=1; i<=N; i++) {
			a[i]=i;
			c[i]=i;
		}
		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]);
			}
		}
		while(q--) {
			int k,x,y,temp;
			scanf("%d%d%d",&k,&x,&y);
			switch(k) {
				case 1: {
					temp=a[x];
					a[x]=a[y];
					a[y]=temp;
					break;
				}
				case 2: {
					temp=c[y];
					c[y]=c[x];
					c[x]=temp;
					break;
				}
				case 3: {
					b[a[x]]+=y;
					break;
				}
				case 4: {
					d[c[x]]+=y;
					break;
				}
			}
		}
		for(int i=1; i<=n; i++) {
			for(int j=1; j<=m; j++) {
				if(j==1)
					printf("%d",map[a[i]][c[j]]+b[a[i]]+d[c[j]]);
				else
					printf(" %d",map[a[i]][c[j]]+b[a[i]]+d[c[j]]);
			}
			printf("\n");
		}
	}
	return 0;
}