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".
题目大意
给出一个 N×N 的矩阵 G，要求判断其是否满足：
- G[i][i]=0；
- G[i][j]=G[j][i]；
- 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。
1 #include<bits/stdc++.h> 2 const int maxn = 2535; 3 const int maxm = 7035; 4 5 struct Edge 6 { 7 int v,val; 8 Edge(int a=0, int b=0):v(a),val(b) {} 9 }edges[maxm]; 10 int n,a[maxn][maxn],f[maxn][maxn]; 11 int edgeTot,head[maxn],nxt[maxm],pre[maxn],dis[maxn]; 12 bool chk,vis[maxn]; 13 14 int read() 15 { 16 char ch = getchar(); 17 int num = 0, fl = 1; 18 for (; !isdigit(ch); ch=getchar()) 19 if (ch=='-') fl = -1; 20 for (; isdigit(ch); ch=getchar()) 21 num = (num<<1)+(num<<3)+ch-48; 22 return num*fl; 23 } 24 void addedge(int u, int v, int c) 25 { 26 edges[++edgeTot] = Edge(v, c), nxt[edgeTot] = head[u], head[u] = edgeTot; 27 edges[++edgeTot] = Edge(u, c), nxt[edgeTot] = head[v], head[v] = edgeTot; 28 } 29 void dfs(int Top, int x, int fa, int c) 30 { 31 f[Top][x] = c; 32 for (int i=head[x]; i!=-1; i=nxt[i]) 33 { 34 int v = edges[i].v; 35 if (v==fa) continue; 36 dfs(Top, v, x, std::max(c, edges[i].val)); 37 } 38 } 39 int main() 40 { 41 memset(head, -1, sizeof head); 42 n = read(); 43 for (int i=1; i<=n; i++) 44 for (int j=1; j<=n; j++) 45 a[i][j] = read(); 46 chk = 1; 47 for (int i=1; i<=n&&chk; i++) 48 if (a[i][i]) chk = false; 49 for (int i=1; i<=n&&chk; i++) 50 for (int j=i+1; j<=n&&chk; j++) 51 if (a[i][j]!=a[j][i]) chk = false; 52 if (!chk) puts("NOT MAGIC"); 53 else{ 54 memset(dis, 0x3f3f3f3f, sizeof dis); 55 dis[1] = 0; 56 for (int T=1; T<=n; T++) 57 { 58 int pos = 0; 59 for (int i=1; i<=n; i++) 60 if (dis[i] < dis[pos]&&!vis[i]) pos = i; 61 if (!pos) break; 62 vis[pos] = true; 63 if (pre[pos]) addedge(pos, pre[pos], dis[pos]); 64 for (int i=1; i<=n; i++) 65 if (!vis[i]&&dis[i] > a[pos][i]) 66 dis[i] = a[pos][i], pre[i] = pos; 67 } 68 for (int i=1; i<=n; i++) dfs(i, i, i, 0); 69 for (int i=1; i<=n&&chk; i++) 70 for (int j=1; j<=n&&chk; j++) 71 if (a[i][j]!=f[i][j]) chk = false; 72 puts(chk?"MAGIC":"NOT MAGIC"); 73 } 74 return 0; 75 }
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