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POJ 1679 The Unique MST（求最小生成树是否唯一）

2024-07-17 01:01:48   224人阅读

The Unique MST
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 20430 Accepted: 7186

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

23 31 2 12 3 23 1 34 41 2 22 3 23 4 24 1 2

Sample Output

3Not Unique!


先求一次最小生成树，然后枚举边，如果去掉这条边的最小生成树价值和原来一样，则不唯一。
 1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <string> 5 #include <iomanip> 6 #include <algorithm> 7 #include <queue> 8 #include <vector> 9 #include <map>10 using namespace std;11 struct Node{12     int u, v, w;13 }node[110*110/2];14 int t, n, m, sum1, sum2,  ans1, ans2;15 int pre[110], mark[110*110/2];16 int find(int x){17     if(pre[x] == x) return x;18     else return find(pre[x]);19 }20 bool cmp(Node x, Node y){21     return x.w < y.w;22 }23 24 int main(){25     scanf("%d", &t);26     while(t--){27         scanf("%d%d", &n, &m);28         for(int i = 1; i <= m; i++){29             scanf("%d%d%d", &node[i].u, &node[i].v, &node[i].w);30         }31         sort(node+1,node+1+m,cmp);32         sum1 = 0;33         ans1 = 0;34         for(int i = 1; i <= n; i++) pre[i] = i;35         for(int i = 1; i <= m; i++){36             if(ans1 == n-1) break;37             int aa = find(node[i].u);38             int bb = find(node[i].v);39             if(aa != bb){40                 sum1 += node[i].w;41                 pre[aa] = bb;42                 mark[++ans1] = i;43             //    ans++;44             }45         }46         bool flag = true;47         for(int k = 1; k <= ans1; k++){48             for(int i = 1; i <= n; i++) pre[i] = i;49             sum2 = 0; ans2 = 0;50             for(int i = 1; i <= m; i++){51                 if(i == mark[k]) continue;52                 if(ans2 == n-1) break;53                 int aa = find(node[i].u);54                 int bb = find(node[i].v);55                 if(aa != bb){56                     sum2 += node[i].w;57                     pre[aa] = bb;58                     ans2++;59                 }60             }61             if(ans2 == n-1 && sum2 == sum1){62                 flag = false; break;63             }64         }65         if(flag) printf("%d\n", sum1);66         else printf("Not Unique!\n");67         68         69 70     }71     72     return 0;73 }

 


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