【每日一题】AcWing3732 矩阵复原【构造】

思路

对于每一行，我们存储每个数所在的列，对于每一列，我们去查询在哪一行，然后存储到一个答案数组即可。

AC代码

C++

#include <iostream>

using namespace std;

const int maxn = 510;
int p[maxn*maxn];
int ans[maxn][maxn];
int main(int argc, char *argv[]) {
	int t;
	cin >> t;
	while(t--){
		int n,m;
		cin >> n >> m;
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= m; j++){
				int x;
				cin >> x;
				p[x] = j;
			}
		}
		for(int i = 1; i <= m; i++){
			int x;
			cin >> x;
			int ind = p[x];
			ans[1][ind] = x;
			for(int j = 2; j <= n; j++){
				cin >> x;
				ans[j][ind] = x;
			}
		}
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= m; j++)
				cout<<ans[i][j]<<" ";
			cout<<endl;
		}
	}
	return 0;
}

Python

t = int(input())
N = 510
p = [-1]*(N**2)
ans = [[-1]*N for _ in range(N)]
for _ in range(t):
    n,m = map(int,input().split())

    # 记录每个数的列号
    for _ in range(n):
        x = list(map(int,input().split()))
        for i in range(1,m+1):
            p[x[i-1]] = i
    for _ in range(m):
        y = list(map(int,input().split()))
        col = p[y[0]]
        for i in range(1,n+1):
            ans[i][col]=y[i-1]
    for i in range(1,n+1):
        for j in range(1,m+1):
            print(ans[i][j],end=" ")
        print("")