POJ 1679 The Unique MST

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37818		Accepted: 13835

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

Sample Output

3
Not Unique!

Source

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdio>
  5 #include <cctype>
  6 
  7 #define int long long
  8 
  9 using namespace std;
 10 
 11 const int N = 405;
 12 
 13 int cge, T, n, m, x, y, z, ec, sum, gc, ans, 
 14 fa[N], f[N], use[N << 9], faz[25][N], mx1[25][N], dep[N];
 15 
 16 struct Edge
 17 {
 18     int u, v, nxt, w;
 19     bool operator < (const Edge &b) const 
 20         { return w < b.w; }
 21 }e[N << 9], g[N << 9];
 22 
 23 inline void read(int &x)
 24 {
 25     int k = 1; x = 0;
 26     char c = getchar();
 27     while (!isdigit(c))
 28         if (c == '-') k = -1, c = getchar();
 29         else c = getchar();
 30     while (isdigit(c))
 31         x = (x << 1) + (x << 3) + (c ^ 48),
 32         c = getchar();
 33     x *= k;
 34 }
 35 
 36 inline void Add(int a, int b, int c)
 37 {
 38     ++gc;
 39     g[gc].u = a;
 40     g[gc].v = b;
 41     g[gc].w = c;
 42 }
 43 
 44 inline void Addedge(int a, int b, int c)
 45 {
 46     ++ec;
 47     e[ec].u = a;
 48     e[ec].v = b;
 49     e[ec].w = c;
 50     e[ec].nxt = f[a];
 51     f[a] = ec;
 52 }
 53 
 54 inline void Init()
 55 {
 56     cge = 9223372036854775807, ec = gc = 0, sum = 0, ans = 0;
 57     read(n), read(m);
 58     memset(f, -1, sizeof(f));
 59     memset(use, 0, sizeof(use));
 60     memset(dep, 0, sizeof(dep));
 61     memset(faz, 0, sizeof(faz));
 62     memset(mx1, 0, sizeof(mx1));
 63     for (int i = 1; i <= n; ++i) fa[i] = i;
 64     for (int i = 1; i <= m; ++i)
 65         read(x), read(y), read(z),
 66         Add(x, y, z);
 67 }
 68 
 69 int Find(int x)
 70 {
 71     return fa[x] == x ? x : fa[x] = Find(fa[x]);
 72 }
 73 
 74 inline bool One(int a, int b)
 75 {
 76     a = Find(a);
 77     b = Find(b);
 78     return a == b;
 79 }
 80 
 81 inline void Union(int a, int b, int u)
 82 {
 83     int ra = Find(a);
 84     int rb = Find(b);
 85     ans += g[u].w; ++sum;
 86     fa[ra] = rb; use[u] = 1;
 87     Addedge(a, b, g[u].w);
 88     Addedge(b, a, g[u].w);
 89     return;
 90 }
 91 
 92 inline void Kruskal()
 93 {
 94     sort(g + 1, g + 1 + gc);
 95     for (int i = 1; i <= gc; ++i)
 96     {
 97         if (!One(g[i].v, g[i].u))
 98             Union(g[i].u, g[i].v, i);
 99         if (sum == n - 1) break;
100     }
101     dep[1] = 1;
102 }
103 
104 void DFS(int u)
105 {
106     for (int i = f[u]; i != -1; i = e[i].nxt)
107     {
108         if (e[i].v == faz[0][u]) continue;
109         faz[0][e[i].v] = u,
110         mx1[0][e[i].v] = e[i].w,
111         dep[e[i].v] = dep[u] + 1,
112         DFS(e[i].v);
113     }
114 }
115 
116 inline void Make()
117 {
118     for (int j = 1; j <= 20; ++j)
119         for (int i = 1; i <= n; ++i)
120             faz[j][i] = faz[j - 1][faz[j - 1][i]],
121             mx1[j][i] = max(mx1[j - 1][i], mx1[j - 1][faz[j - 1][i]]);
122 }
123 
124 inline void LCA(int a, int b, int c)
125 {
126     int m1 = 0, m2 = 0;
127     if (dep[a] < dep[b]) swap(a, b);
128     for (int i = 20; i >= 0; --i)
129         if (dep[faz[i][a]] >= dep[b])
130             m1 = max(m1, mx1[i][a]);
131     if (a == b) { cge = min(cge, c - m1); return; }
132     for (int i = 20; i >= 0; --i)
133         if (faz[i][b] != faz[i][a])
134             m1 = max(m1, max(mx1[i][a], mx1[i][b]));
135     m1 = max(m1, max(mx1[0][a], mx1[0][b]));
136     cge = min(cge, c - m1);
137 }
138 
139 inline void SecMST()
140 {
141     for (int i = 1; i <= m; ++i)
142         if (!use[i]) 
143             LCA(g[i].u, g[i].v, g[i].w);
144     if (cge) printf("%lld
", ans);
145     else puts("Not Unique!");
146 }
147 
148 signed main()
149 {
150     read(T);
151     while (T--) Init(), Kruskal(), DFS(1), Make(), SecMST();
152     return 0;
153 }

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原文地址：https://www.cnblogs.com/yanyiming10243247/p/10127609.html