zoukankan html css js c++ java

Uniqueness of MST

Given any weighted undirected graph, there exists at least one minimum spanning tree (MST) if the graph is connected. Sometimes the MST may not be unique though. Here you are supposed to calculate the minimum total weight of the MST, and also tell if it is unique or not.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers N ( $\leq 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by 3 integers:$

V1 V2 Weight

where V1 and V2 are the two ends of the edge (the vertices are numbered from 1 to N), and Weight is the positive weight on that edge. It is guaranteed that the total weight of the graph will not exceed

Output Specification:

For each test case, first print in a line the total weight of the minimum spanning tree if there exists one, or else print No MST instead. Then if the MST exists, print in the next line Yes if the tree is unique, or No otherwise. There there is no MST, print the number of connected components instead.

Sample Input 1:

Sample Output 1:

11
Yes

Sample Input 2:

Sample Output 2:

4
No

Sample Input 3:

Sample Output 3:

No MST
2

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 vector<vector<pair<int,int> > > v;
  4 vector<vector<pair<int,int> > > vst;
  5 vector<bool> vb;
  6 unordered_set<int> us;
  7 int mst;
  8 struct node
  9 {
 10     int id;
 11     int pid;
 12     int w;
 13     bool operator< (const node& en) const
 14     {
 15         return w>en.w;
 16     }
 17 } no;
 18 vector<node> vsst;
 19 inline int st(int n)
 20 {
 21     priority_queue<node> q;
 22     no.id=n;
 23     no.pid=n;
 24     no.w=0;
 25     q.push(no);
 26     for(;!q.empty();)
 27     {
 28         auto qf=q.top();
 29         q.pop();
 30         if(vb[qf.id])
 31         {
 32             mst=mst+qf.w;
 33             us.emplace(qf.w);
 34             vb[qf.id]=false;
 35             if(qf.id!=qf.pid)
 36             {
 37                 vst[qf.id].emplace_back(make_pair(qf.pid,qf.w));
 38                 vst[qf.pid].emplace_back(make_pair(qf.id,qf.w));
 39             }
 40             for(auto k:v[qf.id])
 41             {
 42                 if(vb[k.first])
 43                 {
 44                     no.id=k.first;
 45                     no.pid=qf.id;
 46                     no.w=k.second;
 47                     q.push(no);
 48                 }
 49             }
 50         }
 51         else
 52         {
 53             if(us.find(qf.w)!=us.end())
 54             vsst.emplace_back(qf);
 55         }
 56     }
 57     for(int i=n;i<vb.size();++i)
 58     if(vb[i])
 59     return i;
 60     return -1;
 61 }
 62 inline bool bfs(int c1,int c2,int w)
 63 {
 64     vector<int> vbb(v.size(),0);
 65     queue<pair<int,int> > q;
 66     q.push(make_pair(c1,1));
 67     q.push(make_pair(c2,-1));
 68     vbb[c1]=1;
 69     vbb[c2]=-1;
 70     for(;!q.empty();)
 71     {
 72         auto qf=q.front();
 73         q.pop();
 74         for(auto k:vst[qf.first])
 75         {
 76             if(vbb[k.first]*qf.second==0)
 77             {
 78                 if(k.second==w)
 79                 qf.second=2*qf.second;
 80                 vbb[k.first]=qf.second;
 81                 q.push(qf);
 82             }
 83             else if(vbb[k.first]*qf.second<0)
 84             {
 85                 if(abs(vbb[k.first])>1)
 86                 return false;
 87                 else
 88                 return true;
 89             }
 90         }
 91     }
 92     return false;
 93 }
 94 int main()
 95 {
 96 //    freopen("data.txt","r",stdin);
 97     int n,m,c1,c2,w;
 98     mst=0;
 99     scanf("%d %d",&n,&m);
100     v.resize(n);
101     vb.resize(n,true);
102     vst.resize(n);
103     for(;m--;)
104     {
105         scanf("%d %d %d",&c1,&c2,&w);
106         c1--;
107         c2--;
108         v[c1].emplace_back(make_pair(c2,w));
109         v[c2].emplace_back(make_pair(c1,w));
110     }
111     w=st(0);
112     if(w<0)
113     {
114         printf("%d
",mst);
115         bool flag=true;
116         for(int i=0;i<vsst.size()&&flag;++i)
117         {
118             if(bfs(vsst[i].id,vsst[i].pid,vsst[i].w))
119             continue;
120             else
121             flag=false;
122         }
123         if(flag)
124         printf("Yes");
125         else
126         printf("No");
127     }
128     else
129     {
130         printf("No MST
");
131         int part=1;
132         for(;w>-1;part++)
133         w=st(w);
134         printf("%d",part);
135     }
136     return 0;
137 }

诚者，君子之所守也。

查看全文

相关阅读:
freeCAD定制界面
 freeCAD预选项编辑器
 freeCAD文档结构
 FreeCAD鼠标操作指南
 freeCAD下载与安装
 freeCAD特性列表
 关于freeCAD
html 试题试卷（包含latex）下载成word
latex转word公式 java （latextoword,latex_word,latex2word,latex_omml）
oracle 行列转换

原文地址：https://www.cnblogs.com/SkystarX/p/12285784.html