次小生成树（poj 1679）

次小生成树

建立在最小生成树的基础上；

由于最小生成树的性质所决定，对于最小生成树任何的两点之间再加一条边，就会形成一个环；

那么，也就是说，加上一条边后，减去一条除刚加上的一条边外权值最大的边，又会构成一棵生成树；

这样，将所有最小生成树上没有的边按上述遍历一次后，得到生成树的最小的权值即为该图的次小生成树。

其中，最大的权值可由prime求最小生成树时顺便求出：

a[x][j]=max{a[x][i], ver[i][j]};

(其中，x为当前在树上的节点，j为当前要入树的点，i为j的父节点)。

代码：

View Code

  1 #include<iostream>
  2 #include<string.h>
  3 using namespace std;
  4 #define INF 1000000000
  5 int first[105];
  6 struct node{
  7        int start;
  8        int end;
  9        int w;
 10        int next;
 11 };
 12 node ver[100005];
 13 int visver[100005];
 14 int a[105];
 15 int vispoint[105];
 16 int secondmax[105][105];
 17 int main(){
 18     int n1;
 19     cin>>n1;
 20     int n,m;
 21     while(n1--){
 22         cin>>n>>m;
 23         memset(first,-1,sizeof(first));
 24         memset(ver,-1,sizeof(ver));
 25         memset(visver,-1,sizeof(visver));
 26         memset(vispoint,-1,sizeof(vispoint));
 27         memset(secondmax,-1,sizeof(secondmax));
 28         for(int i=1;i<=m;i++){
 29                 int temp1,temp2,temp3;
 30                 cin>>temp1>>temp2>>temp3;
 31                 if(first[temp1]==-1){
 32                        first[temp1]=i;
 33                        }
 34                 else{
 35                        int temp=first[temp1];
 36                        first[temp1]=i;
 37                        ver[i].next=temp;
 38                        }
 39                 ver[i].start=temp1;
 40                 ver[i].end=temp2;
 41                 ver[i].w=temp3;
 42                  if(first[temp2]==-1){
 43                        first[temp2]=i+m;
 44                        }
 45                 else{
 46                        int temp=first[temp2];
 47                        first[temp2]=i+m;
 48                        ver[i+m].next=temp;
 49                        }
 50                 ver[i+m].start=temp2;
 51                 ver[i+m].end=temp1;
 52                 ver[i+m].w=temp3;
 53                 }
 54         int times=0;
 55         int start=1;
 56         a[times++]=start;
 57         vispoint[start]=1;
 58         int sum=0;
 59         int flag=0;
 60         for(int j=1;j<n;j++){
 61                 int minn=INF;
 62                 int p=0;
 63                 for(int k=0;k<times;k++){
 64                         for(int g=first[a[k]];g!=-1;g=ver[g].next){
 65                                 if(ver[g].w<minn&&vispoint[ver[g].end]==-1){
 66                                       minn=ver[g].w;
 67                                       p=g;
 68                                       }
 69                                       }
 70                                       }
 71                 if(minn>=INF){
 72                       flag=1;
 73                       break;
 74                       }
 75                 sum+=minn;
 76                 int second=ver[p].w;
 77                 for(int g=0;g<times;g++){
 78                         int temp=second>secondmax[ver[p].start][a[g]]?second:secondmax[ver[p].start][a[g]];
 79                         secondmax[ver[p].end][a[g]]=temp;
 80                         secondmax[a[g]][ver[p].end]=temp;
 81                                 }
 82                 visver[p]=1;
 83                 visver[p+m]=1;
 84                 if(vispoint[ver[p].end]==-1){
 85                      a[times++]=ver[p].end;
 86                      vispoint[ver[p].end]=1;
 87                      }
 88                 }
 89        if(flag==1){
 90             cout<<"unexist SecondMinTree"<<endl;
 91             continue;
 92             }
 93         cout<<sum<<endl;
 94         int sumsec=INF;
 95         for(int j=1;j<=m;j++){
 96                 if(visver[j]==-1){
 97                       if(secondmax[ver[j].start][ver[j].end]==-1)
 98                                 continue;
 99                       int tempsum=sum-secondmax[ver[j].start][ver[j].end];
100                       tempsum+=ver[j].w;
101                       if(tempsum<sumsec)
102                             sumsec=tempsum;
103                             }
104                             }
105        if(sumsec>=INF){
106             cout<<"inexist SecondMinTree"<<endl;
107             continue;
108             }
109        cout<<sumsec<<endl; 
110                        }
111        return 0;
112        }             
113

poj 1679

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15700		Accepted: 5427

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

3
Not Unique!

题意：

图中是否存在相等的最小生成树；有则输出“not Unique”，无则输出最小生成树的值。

View Code

  1 #include<iostream>
  2 #include<string.h>
  3 using namespace std;
  4 #define INF 1000000000
  5 int first[105];
  6 struct node{
  7        int start;
  8        int end;
  9        int w;
 10        int next;
 11 };
 12 node ver[100005];
 13 int visver[100005];
 14 int a[105];
 15 int vispoint[105];
 16 int secondmax[105][105];
 17 int main(){
 18     int n1;
 19     cin>>n1;
 20     int n,m;
 21     while(n1--){
 22         cin>>n>>m;
 23         memset(first,-1,sizeof(first));
 24         memset(ver,-1,sizeof(ver));
 25         memset(visver,-1,sizeof(visver));
 26         memset(vispoint,-1,sizeof(vispoint));
 27         memset(secondmax,-1,sizeof(secondmax));
 28         for(int i=1;i<=m;i++){
 29                 int temp1,temp2,temp3;
 30                 cin>>temp1>>temp2>>temp3;
 31                 if(first[temp1]==-1){
 32                        first[temp1]=i;
 33                        }
 34                 else{
 35                        int temp=first[temp1];
 36                        first[temp1]=i;
 37                        ver[i].next=temp;
 38                        }
 39                 ver[i].start=temp1;
 40                 ver[i].end=temp2;
 41                 ver[i].w=temp3;
 42                  if(first[temp2]==-1){
 43                        first[temp2]=i+m;
 44                        }
 45                 else{
 46                        int temp=first[temp2];
 47                        first[temp2]=i+m;
 48                        ver[i+m].next=temp;
 49                        }
 50                 ver[i+m].start=temp2;
 51                 ver[i+m].end=temp1;
 52                 ver[i+m].w=temp3;
 53                 }
 54         int times=0;
 55         int start=1;
 56         a[times++]=start;
 57         vispoint[start]=1;
 58         int sum=0;
 59         int flag=0;
 60         for(int j=1;j<n;j++){
 61                 int minn=INF;
 62                 int p=0;
 63                 for(int k=0;k<times;k++){
 64                         for(int g=first[a[k]];g!=-1;g=ver[g].next){
 65                                 if(ver[g].w<minn&&vispoint[ver[g].end]==-1){
 66                                       minn=ver[g].w;
 67                                       p=g;
 68                                       }
 69                                       }
 70                                       }
 71                 if(minn>=INF){
 72                       flag=1;
 73                       break;
 74                       }
 75                 sum+=minn;
 76                 int second=ver[p].w;
 77         //        cout<<p<<" "<<ver[p].start<<" "<<ver[p].end<<" "<<ver[p].w<<endl; 
 78                 for(int g=0;g<times;g++){
 79                         int temp=second>secondmax[ver[p].start][a[g]]?second:secondmax[ver[p].start][a[g]];
 80           //              cout<<ver[p].end<<" "<<a[times]<<endl;
 81                         secondmax[ver[p].end][a[g]]=temp;
 82                         secondmax[a[g]][ver[p].end]=temp;
 83                   //      cout<<temp<<" "<<secondmax[ver[p].start][a[times]]<<endl;
 84                                 }
 85                 visver[p]=1;
 86                 visver[p+m]=1;
 87                 if(vispoint[ver[p].end]==-1){
 88                      a[times++]=ver[p].end;
 89                      vispoint[ver[p].end]=1;
 90                      }
 91                 }
 92        if(flag==1){
 93             cout<<"0"<<endl;
 94           //  cout<<"unexist SecondMinTree"<<endl;
 95             continue;
 96             }
 97     //    cout<<sum<<endl;
 98  /*       for(int i=1;i<=n;i++){
 99                 for(int k=1;k<=n;k++){
100                         cout<<i<<" "<<k<<" "<<secondmax[i][k]<<endl;
101                         }
102                         }*/
103         int sumsec=INF;
104         for(int j=1;j<=m;j++){
105                 if(visver[j]==-1){
106                       if(secondmax[ver[j].start][ver[j].end]==-1)
107                                 continue;
108                       int tempsum=sum-secondmax[ver[j].start][ver[j].end];
109                       tempsum+=ver[j].w;
110                       if(tempsum<sumsec)
111                             sumsec=tempsum;
112                             }
113                             }
114    /*    if(sumsec>=INF){
115             cout<<"inexist SecondMinTree"<<endl;
116             continue;
117             }*/
118        if(sum==sumsec){
119            cout<<"Not Unique!"<<endl;
120            }
121        else
122            cout<<sum<<endl; 
123                        }
124        return 0;
125        }             
126

posted on 2012-11-14 19:19 yumao 阅读(240) 评论(0) 收藏举报

刷新页面返回顶部

yumao