题目连接：

http://www.codeforces.com/contest/632/problem/F

Description

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".

Sample Input

3
0 1 2
1 0 2
2 2 0

Sample Output

MAGIC

Hint

题意

给你一个nn的矩阵，然后判断是否是magic的

如何是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：

就正解是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]

于是就这样做呗。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2505;
int a[maxn][maxn];
pair<int,pair<int,int> > d[maxn*maxn];
int n,tot,last;
int fa[maxn];
int fi(int x)
{return x == fa[x]?x:fa[x]=fi(fa[x]);
}
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&a[i][j]);for(int i=1;i<=n;i++){fa[i]=i;for(int j=1;j<=n;j++){if(i==j&&a[i][j])return puts("NOT MAGIC");if(a[i][j]!=a[j][i])return puts("NOT MAGIC");if(i>j)d[tot++]=make_pair(a[i][j],make_pair(i,j));}}sort(d,d+tot);for(int i=0;i<tot;i++){if(i+1<tot&&d[i].first==d[i+1].first)continue;for(int j=last;j<=i;j++){int p1 = fi(d[j].second.first);int p2 = fi(d[j].second.second);if(p1==p2)return puts("NOT MAGIC");}for(int j=last;j<=i;j++){int p1 = fi(d[j].second.first);int p2 = fi(d[j].second.second);fa[p2]=p1;}last=i+1;}puts("MAGIC");
}