POJ1679:The Unique MST(最小生成树)
作者:互联网
The Unique MST
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 38430 | Accepted: 14045 |
题目链接: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!
题意:
判断最小生成树是否唯一,如果是就输出最小生成树的边权和。
题解:
对于权值相同的边,先把不能加入的边去除掉,然后把能加的边都加进图中,如果还剩有边,那么说明最小生成树不是唯一的。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> using namespace std; typedef long long ll; const int N = 105; int t,n,m; struct Edge{ int u,v,w; bool operator < (const Edge &A)const{ return w<A.w; } }e[N*N]; int f[N]; int find(int x){ return f[x]==x?f[x]:f[x]=find(f[x]); } int Kruskal(){ int ans=0,cnt,j; for(int i=0;i<=n+1;i++) f[i]=i; for(int i=1;i<=m;i++){ j=i;cnt=0; while(e[i].w==e[j].w && j<=m) j++,cnt++; for(int k=i;k<j;k++){ int fx=find(e[k].u),fy=find(e[k].v); if(fx==fy) cnt--; } for(int k=i;k<j;k++){ int fx=find(e[k].u),fy=find(e[k].v); if(fx!=fy){ f[fx]=fy; ans+=e[i].w; cnt--; } } if(cnt>0) return -1; } return ans ; } int main(){ cin>>t; while(t--){ scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); e[i].u=u;e[i].v=v;e[i].w=w; } sort(e+1,e+m+1); int ans = Kruskal(); if(ans==-1) puts("Not Unique!"); else printf("%d\n",ans); } return 0; }
标签:Unique,POJ1679,int,MST,tree,connected,spanning,include 来源: https://www.cnblogs.com/heyuhhh/p/10372197.html