描述

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'.

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

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

樣例輸出

3
Not Unique!

來源

POJ Monthly--2004.06.27 [email protected]

 1 #include <iostream>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <stack>
 5 #include <string>
 6 #include <math.h>
 7
 
 #include <queue>
 8 #include <stdio.h>
 9 #include <string.h>
10 #include <set>
11 #include <vector>
12 #include <fstream>
13 #define maxn 10005
14 #define inf 999999
15 #define cha 127
16 using namespace std;
17 
18 struct node {
19     int x, y, weight;
20     bool valid;
21     node() {
22         valid = true;
23     }
24 }all[maxn];
25 bool operator<(node a, node b) {
26     return a.weight < b.weight;
27 }
28 int p[105], n, m;
29 
30 int parent(int x) {
31     if (p[x] == x)return x;
32     p[x] = parent(p[x]);
33     return p[x];
34 }
35 
36 void merge(int x, int y) {
37     int px = parent(x), py = parent(y);
38     p[px] = py;
39 }
40 
41 vector<int>q;
42 
43 int MST(int ifq) {
44     int ans=0,num=n;
45     for (int i = 1; i <= n; i++)
46         p[i] = i;
47     for (int i = 1; i <= m; i++) {
48         int x = all[i].x, y = all[i].y, val = all[i].weight;
49         if (all[i].valid&&parent(x) != parent(y)) {
50             merge(x, y);
51             ans += val;
52             num--;
53             if (ifq)
54                 q.push_back(i);
55         }
56         if (num==1)
57             break;
58     }
59     if (num != 1)
60         return -1;
61     return ans;
62 }
63 
64 void solve() {
65     int ans = MST(1), size = q.size();
66     for (int i = 0; i < size; i++) {
67         all[q[i]].valid = false;
68         int tmp=MST(0);
69         all[q[i]].valid = true;
70         if (tmp == ans) {
71             printf("Not Unique!\n");
72             return;
73         }
74     }
75     printf("%d\n", ans);
76 }
77 
78 void init() {
79     q.clear();
80     scanf("%d%d", &n, &m);
81     for (int i = 1; i <= m; i++) 
82         scanf("%d%d%d", &all[i].x, &all[i].y, &all[i].weight);
83     sort(all + 1, all + 1 + m);
84     solve();
85 }
86 
87 int main() {
88     int kase;
89     scanf("%d", &kase);
90     while (kase--)
91         init();
92     return 0;
93 }

wa點：

1)演算法不能寫錯……

2)MST的時候要考慮刪邊後不連通的情況

3)每次初始化要把上次遺留的資料刪乾淨

做完模擬題感到了空虛和無奈……