POJ 1679 The Unique MST(次小生成树&Kruskal)
2013-08-06 23:46
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The Unique MST
http://poj.org/problem?id=1679
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
思路:
先求一个MST,再从原图G中删掉从有向树(以找的次序为方向)的每个点出发的边(每次删掉一条边),对删掉一条边的新图G'求MST与原有比较即可。
完整代码:
http://poj.org/problem?id=1679
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!
思路:
先求一个MST,再从原图G中删掉从有向树(以找的次序为方向)的每个点出发的边(每次删掉一条边),对删掉一条边的新图G'求MST与原有比较即可。
完整代码:
/*16ms,224KB*/ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int Max_V = 102; const int Max_E = 4952; struct EDGE { int u, v, w; bool ischoosed; } edge[Max_E]; int path[Max_V], V, E, bj[Max_V + 2]; bool cmp(const EDGE&a, const EDGE&b) { return a.w < b.w; } int search_path(int x) { if (x != path[x]) path[x] = search_path(path[x]); return path[x]; } int Kruskal(int start) { int count = 0, sum = 0, x, y; for (int i = 1; i <= V; i++) path[i] = i; for (int i = start; i < E && count < V - 1; i++) { x = search_path(edge[i].u); y = search_path(edge[i].v); if (x == y) continue; bj[count++] = i;//?????????????????????? path[x] = path[y];//修正 sum += edge[i].w; } return count < V - 1 ? 0 : sum; } int Seckruskal() { int count = 0, sum = 0, x, y; for (int i = 1; i <= V; i++) path[i] = i; for (int i = 0; i < E && count < V - 1; i++) { x = search_path(edge[i].u); y = search_path(edge[i].v); if (x == y || edge[i].ischoosed == true)//把这条边删掉,看有没有另外的MST continue; count++; path[x] = path[y]; sum += edge[i].w; } return count < V - 1 ? 0 : sum; } int main() { int T, i, temp, minMST; bool flag; scanf("%d", &T); while (T--) { memset(bj, 0, sizeof(bj)); memset(path, 0, sizeof(path)); memset(edge, 0, sizeof(edge)); flag = false; scanf("%d%d", &V, &E); for (i = 0; i < E; i++) scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w); sort(edge, edge + E, cmp); /////////////////////////////////////////////////////// minMST = Kruskal(0); for (i = 0; i < V - 1; i++) { edge[bj[i]].ischoosed = true; temp = Seckruskal(); edge[bj[i]].ischoosed = false; if (temp == minMST && temp) { flag = true; break; } } if (flag) printf("Not Unique!\n"); else printf("%d\n", minMST); } return 0; }
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