【hdoj5671】Matrix

分手后的思念是犯贱 2022-09-25 15:16 112阅读 0赞

# 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)][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;
    }

[BestCoder Round _81 _div.2]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2381+%28div.2%29&source=1&searchmode=source