（Educational Codeforces Round 9）Magic Matrix（最小生成树）

（Educational Codeforces Round 9）Magic Matrix（最小生成树）

Magic Matrix

time limit per test5 seconds memory limit per test512 megabytes inputstandard input outputstandard output You’re given a matrix A of size n × n.

Let’s call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all triples i, j, k. Note that i, j, k do not need to be distinct.

Determine if the matrix is magic.

As the input/output can reach very huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 2500) — the size of the matrix A.

Each of the next n lines contains n integers aij (0 ≤ aij < 109) — the elements of the matrix A.

Note that the given matrix not necessarily is symmetric and can be arbitrary.

Output

Print ”MAGIC” (without quotes) if the given matrix A is magic. Otherwise print ”NOT MAGIC”.

Examples

input

3 0 1 2 1 0 2 2 2 0

output

MAGIC

input

2 0 1 2 3

output

NOT MAGIC

input

4 0 1 2 3 1 0 3 4 2 3 0 5 3 4 5 0

output

NOT MAGIC

题意

给你一个n*n的矩阵，然后判断是否是magic的 什么是magic的呢？只要a[i][j]=a[j][i],a[i][i]=0,a[i][j]<=max(a[i][k],a[k][j])对于所有的k。

题解： 把这个矩阵抽象成一个完全图 然后这个完全图，a[i][j]表示i点向j点连一条a[i][j]的边 然后magic是什么意思呢？ 就是任何一条路径中的最大边都大于等于a[i][j]，那么翻译过来就是跑最小生成树的之后，i到j上的最大边大于等于a[i][j]。

#include #define x first#define y secondusing namespace std;const int N = 2510;int a[N][N];vector > v;bitset b[N];int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { scanf("%d", &a[i][j]); v.push_back(make_pair(a[i][j], i * n + j)); } if (a[i][i] != 0){ printf("NOT MAGIC\n"); return 0; } } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (a[i][j] != a[j][i]) { printf("NOT MAGIC\n"); return 0; } } } sort(v.begin(), v.end()); int l = 0; for (int k = 0; k < v.size(); k++) { int i = v[k].y / n, j = v[k].y % n; if ((b[i] & b[j]).any()) { printf("NOT MAGIC\n"); return 0; } if (k != v.size() - 1 && v[k].x != v[k + 1].x) { for (; l <= k; l++){ b[v[l].y / n][v[l].y % n] = 1; b[v[l].y / n][v[l].y % n] = 1; } } } printf("MAGIC\n");}

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