poj1679 The Unique MST
2016-09-21 17:17
337 查看
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!’.
先求最小生成树,然后枚举换掉某一条边,看是否仍和原来的大小相等。
就具体做法就是枚举强制不选每一条边,然后求最小生成树。
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!’.
先求最小生成树,然后枚举换掉某一条边,看是否仍和原来的大小相等。
就具体做法就是枚举强制不选每一条边,然后求最小生成树。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct edge { int f,t,l; bool operator < (const edge & ee) const { return l<ee.l; } }g[10010]; int n,m,fa[110],use[10010],ban[10010]; bool mark; int find(int x) { return x==fa[x]?x:fa[x]=find(fa[x]); } int kru() { int i,j,k,x,y,z,ans=0,cnt=0; for (i=1;i<=n;i++) fa[i]=i; for (i=1;cnt<n-1;i++) if (!ban[i]) { x=find(g[i].f); y=find(g[i].t); if (x==y) continue; ans+=g[i].l; cnt++; fa[x]=y; if (mark) use[i]=1; } return ans; } int main() { int i,j,k,x,y,z,T,ans; bool flag; scanf("%d",&T); while (T--) { scanf("%d%d",&n,&m); for (i=1;i<=m;i++) use[i]=ban[i]=0; for (i=1;i<=m;i++) scanf("%d%d%d",&g[i].f,&g[i].t,&g[i].l); sort(g+1,g+m+1); mark=1; ans=kru(); mark=0; flag=0; for (i=1;i<=m;i++) if (use[i]) { ban[i]=1; if (ans==kru()) { flag=1; break; } ban[i]=0; } if (flag) printf("Not Unique!\n"); else printf("%d\n",ans); } }
相关文章推荐
- POJ-1679 The Unique MST 次小生成树
- poj1679——The Unique MST
- POJ 1679 The Unique MST(次小生成树)
- poj 1679 The Unique MST (次小生成树)
- POJ 1679 The Unique MST
- POJ 1679 The Unique MST
- poj_1679 The Unique MST
- POJ 1679 The Unique MST 次小生成树Prim
- poj 1679 The Unique MST (最小生成树)
- poj1679——The Unique MST
- poj 1679 The Unique MST(最小树不唯一的判定)
- poj 1679 The Unique MST
- poj 1679 The Unique MST ( 次小生成树 )
- POJ 1679 The Unique MST (次小生成树Prime/Kruskal)
- poj 1679 The Unique MST(次小生成树)
- poj 1679 The Unique MST
- POJ1679:The Unique MST
- POJ 1679 The Unique MST(判断最小生成树是否唯一)
- POJ 1679 The Unique MST
- poj 1679 The Unique MST(次小生成树)