【思维题 经典模型】cf632F. Magic Matrix

非常妙的经典模型转化啊……

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.outin 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".

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题目大意

给出一个 N×N 的矩阵 G，要求判断其是否满足：

1. G[i][i]=0；
2. G[i][j]=G[j][i]；
3. 对于所有的 i，j，k，有 G[i][j]≤max（G[i][k]，G[j][k]）。

题目分析

这个矩阵的形式非常特殊，然而考场上只是一直在想$n^{2.66}$的鬼畜矩乘……

本题是经典模型的巧妙转化。这里给出的矩阵实际上是一个邻接矩阵的形式，而第三个限制条件说的就是：原图最小瓶颈生成树的距离矩阵等于原矩阵。

 #include<bits/stdc++.h>
 const int maxn = ;
 const int maxm = ;

 struct Edge
 {
     int v,val;
     Edge(int a=, int b=):v(a),val(b) {}
 }edges[maxm];
 int n,a[maxn][maxn],f[maxn][maxn];
 int edgeTot,head[maxn],nxt[maxm],pre[maxn],dis[maxn];
 bool chk,vis[maxn];

 int read()
 {
     char ch = getchar();
     int num = , fl = ;
     for (; !isdigit(ch); ch=getchar())
         if (ch=='-') fl = -;
     for (; isdigit(ch); ch=getchar())
         num = (num<<)+(num<<)+ch-;
     return num*fl;
 }
 void addedge(int u, int v, int c)
 {
     edges[++edgeTot] = Edge(v, c), nxt[edgeTot] = head[u], head[u] = edgeTot;
     edges[++edgeTot] = Edge(u, c), nxt[edgeTot] = head[v], head[v] = edgeTot;
 }
 void dfs(int Top, int x, int fa, int c)
 {
     f[Top][x] = c;
     for (int i=head[x]; i!=-; i=nxt[i])
     {
         int v = edges[i].v;
         if (v==fa) continue;
         dfs(Top, v, x, std::max(c, edges[i].val));
     }
 }
 int main()
 {
     memset(head, -, sizeof head);
     n = read();
     for (int i=; i<=n; i++)
         for (int j=; j<=n; j++)
             a[i][j] = read();
     chk = ;
     for (int i=; i<=n&&chk; i++)
         if (a[i][i]) chk = false;
     for (int i=; i<=n&&chk; i++)
         for (int j=i+; j<=n&&chk; j++)
             if (a[i][j]!=a[j][i]) chk = false;
     if (!chk) puts("NOT MAGIC");
     else{
         memset(dis, 0x3f3f3f3f, sizeof dis);
         dis[] = ;
         for (int T=; T<=n; T++)
         {
             int pos = ;
             for (int i=; i<=n; i++)
                 if (dis[i] < dis[pos]&&!vis[i]) pos = i;
             if (!pos) break;
             vis[pos] = true;
             if (pre[pos]) addedge(pos, pre[pos], dis[pos]);
             for (int i=; i<=n; i++)
                 if (!vis[i]&&dis[i] > a[pos][i])
                     dis[i] = a[pos][i], pre[i] = pos;
         }
         for (int i=; i<=n; i++) dfs(i, i, i, );
         for (int i=; i<=n&&chk; i++)
             for (int j=; j<=n&&chk; j++)
                 if (a[i][j]!=f[i][j]) chk = false;
         puts(chk?"MAGIC":"NOT MAGIC");
     }
     return ;
 }

END