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!
给你一张图，问你能不能构造出两棵不同的最小生成树，方法是先构造出一个最小生成树，然后枚举该生成树的某一条边不能用，再次构造生成树，比较生成树权值。
#include<iostream>#include<string.h>#include<stdio.h>#include<string>#include<vector>#include<algorithm>using namespace std;int n,m;struct edge{ int beg, endd, len; edge(){} edge(int a, int b, int c){ beg = a, endd = b, len = c; }}e[5005];bool cmp(edge a,edge b){ return a.len < b.len;}int f[105];int F(int x){ if (f[x] == x)return x; return f[x] = F(f[x]);}vector<int> ansn;int main(){ int x, y,z,t; scanf("%d", &t); while (t--){ ansn.clear(); scanf("%d%d", &n, &m); for (int i = 0; i < m; i++){ scanf("%d%d%d", &x, &y, &z); e[i] = edge(x, y, z); } sort(e, e + m, cmp); int ans = 0; for (int i = 1; i <= n; i++){ f[i] = i; } for (int i = 0; i < m; i++){ int a = e[i].beg, b = e[i].endd, c = e[i].len; if (F(a) != F(b)){ ansn.push_back(i); ans += c; f[F(a)] = F(b); } } bool same = 0; int total = ansn.size(); for (int i = 0; i < total; i++){ int cntans = 0,cntto=0; for (int k = 1; k <= n; k++){ f[k] = k; } for (int j = 0; j < m; j++){ if (j != ansn[i]){ int a = e[j].beg, b = e[j].endd, c = e[j].len; if (F(a) != F(b)){ cntans += c; cntto++; f[F(a)] = F(b); } } } if (cntto == total&&cntans == ans){ same = 1; break; } } if (same){ printf("Not Unique!/n"); } else{ printf("%d/n", ans); } } return 0;}新闻热点
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