poj1679 次最小生成树 kruskal（暴力枚举)

最新推荐文章于 2019-01-29 20:53:13 发布

转载最新推荐文章于 2019-01-29 20:53:13 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/qldabiaoge/p/9080058.html

本文介绍了一种通过暴力枚举的方法来判断给定图的最小生成树是否唯一的算法实现。该方法首先找到一棵最小生成树，并记录其总权重，然后逐一删除已使用过的边，尝试寻找另一棵总权重相同的生成树。

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!

这题直接暴力枚举，找到一颗最小生成树，标记一下，依次删边

 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 }

转载于:https://www.cnblogs.com/qldabiaoge/p/9080058.html