hdu-5671 Matrix
题目链接:
Matrix
Time Limit: 3000/1500 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
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);
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.
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
题意:
一个矩阵交换行或者列,或者把某一行或者某一列加一个数;
思路:
用一个数组记录位置,一个记录加的数值;
这题有个奇妙的TLE点就是我开的的t,x,y为全局变量时会TLE,开成局部的就过了,求懂原理的大神留言讲解;
AC代码:
#include <bits/stdc++.h> using namespace std; typedef long long ll; const ll mod=1e9+7; const ll inf=1e15; const int N=1006; int a[N][N],l[N],r[N],fl[N],fr[N]; int main() { int t; scanf("%d",&t); while(t--) { int n,m,q; scanf("%d%d%d",&n,&m,&q); for(int i=1;i<=n;i++)l[i]=i,fl[i]=0; for(int i=1;i<=m;i++)r[i]=i,fr[i]=0; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]); int t,x,y; for(int i=1;i<=q;i++) { scanf("%d%d%d",&t,&x,&y); if(t==1) swap(l[x],l[y]); else if(t==2) swap(r[x],r[y]); else if(t==3) fl[l[x]]+=y; else fr[r[x]]+=y; } for(int i=1;i<=n;i++) { for(int j=1;j<m;j++) { printf("%d ",a[l[i]][r[j]]+fl[l[i]]+fr[r[j]]); } printf("%d\n",a[l[i]][r[m]]+fl[l[i]]+fr[r[m]]); } } return 0; }