BC #53 1002 Rikka with Tree

谁借莪１个温暖的怀抱￠ 2021-12-20 11:27 192阅读 0赞

# Rikka with Tree #

Accepts: 207

Submissions: 815

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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:

For a tree TT, let F(T,i)F(T,i) be the distance between vertice 1 and vertice ii.(The length of each edge is 1).

Two trees AA and BB are similiar if and only if the have same number of vertices and for each ii meet F(A,i)=F(B,i)F(A,i)=F(B,i).

Two trees AA and BB are different if and only if they have different numbers of vertices or there exist an number ii which vertice ii have different fathers in tree AA and tree BB when vertice 1 is root.

Tree AA is special if and only if there doesn't exist an tree BB which AA and BB are different and AA and BB are similiar.

Now he wants to know if a tree is special.

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

Input

There are no more than 100 testcases.

For each testcase, the first line contains a number n(1 \\leq n \\leq 1000)n(1≤n≤1000).

Then n-1n−1 lines follow. Each line contains two numbers u,v(1 \\leq u,v \\leq n)u,v(1≤u,v≤n) , which means there is an edge between uu and vv.

Output

For each testcase, if the tree is special print "YES" , otherwise print "NO".

Sample Input

3
    1 2
    2 3
    4
    1 2
    2 3
    1 4

Sample Output

YES
    NO

Hint

For the second testcase, this tree is similiar with the given tree: 4 1 2 1 4 3 4

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

1 #include <stdio.h>
     2 #include <string.h>
     3 #include <vector>
     4 #include <algorithm>
     5 using namespace std;
     6 vector <int> edg[1005];
     7 int s[1005],q[1005],n,m;
     8 int d(int x,int y)
     9 {
    10     int i,j,k;
    11     for(i=0;i<edg[x].size();i++)
    12     {
    13         int v=edg[x][i];
    14         if(s[v]==0)
    15         {
    16             s[v]=y+1;
    17             d(v,y+1);
    18             if(y+1>m)
    19                 m=y+1;
    20         }
    21     }
    22 }
    23 int main()
    24 {
    25     int i,j,k;
    26     while(scanf("%d",&n)!=EOF)
    27     {
    28         for(i=1;i<=n;i++)
    29         {
    30             if(edg[i].size())
    31             {
    32                 edg[i].clear();
    33             }
    34         }
    35         memset(s,0,sizeof(s));
    36         memset(q,0,sizeof(q));
    37         for(i=1;i<n;i++)
    38         {
    39             int x,y;
    40             scanf("%d %d",&x,&y);
    41             edg[x].push_back(y);
    42             edg[y].push_back(x);
    43         }
    44         s[1]=1;m=0;
    45         d(1,1);
    46         for(i=1;i<=n;i++)
    47             q[s[i]]++;
    48         int flg=0;
    49         for(i=1;i<m;i++)
    50         {
    51             if(q[i]>1)
    52             {
    53                 flg=1;
    54                 break;
    55             }
    56         }
    57         if(flg==1)
    58             printf("NO\n");
    59         else
    60             printf("YES\n");
    61     }
    62     return 0;
    63 }

转载于:https://www.cnblogs.com/cyd308/p/4770793.html

[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
[ExpandedBlockStart.gif]: /images/20211220/dd14c8002f0d4bf8a665ef4d2e103b8e.png