【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

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 
 

 

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]]);elseprintf(" %d",map[a[i]][c[j]]+b[a[i]]+d[c[j]]);}printf("\n");}}return 0;}