them tel represent prope tin 生成樹 def use AR

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

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

Sample Output

3
Not Unique!

這題直接暴力枚舉，找到一顆最小生成樹，標記一下，依次刪邊

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4
 
 #include <stack>
 5 #include <string>
 6 #include <math.h>
 7 using namespace std;
 8 const int maxn = 1010 ;
 9 const int INF = 0x7fffffff;
10 struct node {
11     int u, v, w;
12     int used, num, flag;
13 } qu[10 * maxn];
14 int fa[maxn], n, m;
15 int cmp(node a, node b) {
16     return a.w < b.w;
17 }
18 void init() {
19     for (int i = 0 ; i <= n ; i++) fa[i] = i;
20 }
21 int Find(int x) {
22     return fa[x] == x ? x : fa[x] = Find(fa[x]);
23 }
24 int combine(int x, int y) {
25     int nx = Find(x);
26     int ny = Find(y);
27     if (nx != ny) {
28         fa[nx] = ny;
29         return 1;
30     }
31     return 0;
32 }
33 int kruskal(int flag) {
34     init();
35     int sum = 0, cnt = 0;
36     for (int i = 0 ; i < m ; i++) {
37         if (qu[i].flag) continue;
38         if (combine(qu[i].v, qu[i].u)) {
39             if (!flag) qu[i].used = 1;
40             sum += qu[i].w;
41             cnt++;
42             if (cnt == n - 1) break;
43         }
44     }
45     if (cnt != n - 1) return -1;
46     return sum;
47 }
48 int main() {
49     int t;
50     scanf("%d", &t);
51     while(t--) {
52         scanf("%d%d", &n, &m);
53         for (int  i = 0 ; i < m ; i++) {
54             scanf("%d%d%d", &qu[i].u, &qu[i].v, &qu[i].w);
55             qu[i].flag = 0, qu[i].used = 0;
56         }
57         sort(qu, qu + m, cmp);
58         int sum = kruskal(0);
59         int flag = 0;
60         for (int i = 0 ; i < m ; i++) {
61             if (qu[i].used ) {
62                 qu[i].flag = 1;
63                 int temp = kruskal(1);
64                 qu[i].flag = 0;
65                 if (temp == sum) {
66                     flag = 1;
67                     break;
68                 }
69             }
70         }
71         if (flag) printf("Not Unique!\n");
72         else printf("%d\n", sum);
73     }
74     return 0;
75 }

poj1679 次最小生成樹 kruskal（暴力枚舉)