图论（生成树）：HDU 5631Rikka with Graph

Posted 2020-06-14

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 118    Accepted Submission(s): 52

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.

Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.

It is too difficult for Rikka. Can you help her?

 

 

Input

The first line contains a number T(T≤30)——The number of the testcases.

For each testcase, the first line contains a number n(n≤100).

Then n+1 lines follow. Each line contains two numbers u,v , which means there is an edge between u and v.

 

 

Output

For each testcase, print a single number.

 

 

Sample Input

1 3 1 2 2 3 3 1 1 3

 

 

Sample Output

9

 

 

Source

BestCoder Round #73 (div.2)

 

 

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hujie   |   We have carefully selected several similar problems for you:  5634 5633 5632 5630 5629 

 

　　题意：给出一张 n 个点 n+1 条边的无向图，你可以选择一些边（至少一条）删除。有多少种方案使得删除之后图依然联通。

　　这是best coder上面比赛的题目，可我比赛时数组开小了！！！！！

　　内心哔了狗有木有！！！！！！！！！！！！！！！！！！！！！

　　这里我用了组合数学，0ms过（其实暴力N³也可以过）~~~

　　第一名哦！！！

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 using namespace std;
  5 const int maxn=110;
  6 int edge[maxn][3];
  7 int fa[maxn],cnt=1,fir[maxn],nxt[maxn<<1],to[maxn<<1];
  8 int dep[maxn],pre[maxn],ret[maxn],ID[maxn<<1];
  9 int find(int x)
 10 {
 11     return fa[x]==x?x:fa[x]=find(fa[x]);
 12 }
 13 
 14 void addedge(int a,int b,int id)
 15 {
 16     nxt[++cnt]=fir[a];ID[cnt]=id;
 17     to[cnt]=b;fir[a]=cnt;
 18 }
 19 
 20 void DFS(int node)
 21 {
 22     for(int i=fir[node];i;i=nxt[i])
 23     {
 24         if(dep[to[i]])continue;
 25         dep[to[i]]=dep[node]+1;
 26         pre[to[i]]=i;
 27         DFS(to[i]);
 28     }
 29 }
 30 int Search(int x,int y,int d)
 31 {
 32     while(x!=y){
 33         if(dep[x]<dep[y])
 34             swap(x,y);
 35         ret[ID[pre[x]]]+=d;
 36         x=to[pre[x]^1];
 37     }
 38 }
 39 void Init()
 40 {
 41     memset(fir,0,sizeof(fir));
 42     memset(ret,0,sizeof(ret));
 43     cnt=1;
 44 }
 45 
 46 int main()
 47 {    
 48     freopen("data.in","r",stdin);
 49     freopen("wrong.out","w",stdout);
 50     int T,n;
 51     scanf("%d",&T);
 52     while(T--)
 53     {
 54         Init();
 55         scanf("%d",&n);
 56         for(int i=1;i<=n+1;i++){
 57             fa[i]=i;
 58         }
 59     
 60         for(int i=1;i<=n+1;i++)
 61             scanf("%d%d",&edge[i][0],&edge[i][1]);
 62         for(int i=1;i<=n+1;i++){
 63             int a=find(edge[i][0]),b=find(edge[i][1]);
 64             if(a==b)
 65                 edge[i][2]=1;
 66             else{
 67                 fa[a]=b;
 68                 addedge(edge[i][0],edge[i][1],i);
 69                 addedge(edge[i][1],edge[i][0],i);
 70             }
 71         }
 72         int flag=0;
 73         for(int i=2;i<=n;i++)
 74         if(find(i)!=find(i-1)){
 75             printf("0\n");
 76             flag=1;
 77             break;
 78         }
 79         if(flag==1)continue;
 80         dep[1]=1;
 81         DFS(1);
 82         int ans=0,a=0,b=0,c=0;
 83         for(int i=1;i<=n+1;i++){
 84             if(edge[i][2]){
 85                 if(!ans){
 86                     Search(edge[i][0],edge[i][1],1);
 87                     ret[i]+=1;
 88                 }
 89                     
 90                 else {
 91                     Search(edge[i][0],edge[i][1],10);
 92                     ret[i]+=10;
 93                 }
 94                 ans+=1;    
 95             }
 96         }
 97         for(int i=1;i<=n+1;i++){
 98             if(ret[i]==1)a++;
 99             else if(ret[i]==10)b++;
100             else if(ret[i]==11)c++,b++,a++;
101         }
102         printf("%d\n",a+b-c+(a+b-c-1)*(a+b-c)/2-(a-c)*(a-c-1)/2-(b-c)*(b-c-1)/2-c*(c-1)/2);
103     }
104     return 0;
105 }