HDU 5423 Rikka with Tree(水题)

Rikka with Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 292 Accepted Submission(s): 149



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

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

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

Tree A is
special if and only if there doesn't exist an tree B which A and B are
different and A and B 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≤n≤1000).

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



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

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

Source

BestCoder Round #53 (div.2)

中文题意：

问题描述

众所周知，萌萌哒六花不擅长数学，所以勇太给了她一些数学问题做练习，其中有一道是这样的：

对于一棵树TT,令F(T,i)F(T,i)为点1到点ii的最短距离（边长是1）. 

两棵树AA和BB是相似的当且仅当他们顶点数相同且对于任意的ii都有F(A,i)=F(B,i)F(A,i)=F(B,i).

两棵树AA和BB是不同的当且仅当他们定点数不同或者存在一个ii使得以1号点为根的时候ii在两棵树中的父亲不同。

一棵树AA是特殊的当且仅当不存在一棵和它不同的树和它相似。

现在勇太想知道一棵树到底是不是特殊的。

当然，这个问题对于萌萌哒六花来说实在是太难了，你可以帮帮她吗？

输入描述

数据组数不超过100组。每组数据的第一行一个整数n(2 \leq n \leq 1000)n(2≤n≤1000)。

接下来n-1n−1行。每行两个整数u,v(1 \leq u,v \leq n)u,v(1≤u,v≤n)，代表给定树上的一条边。

输出描述

对于每一组数据，如果给定树是特殊的输出"YES"否则输出"NO"。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
int v[1010];
int map[1001][1001];

int main() {
    while(scanf("%d",&n)!=EOF) {
        memset(map,0,sizeof(map));
        memset(v,0,sizeof(v));
        int x,y;
        int pf;
        for(int i=0; i<n-1; i++) {
            scanf("%d%d",&x,&y);
            map[x][y] = 1;
            map[y][x] = 1;
            if(x == 1) {
                pf = y;
            } else if(y == 1) {
                pf = x;
            }
            v[x]++;
            v[y]++;
        }
        int rt = 0;
        int cnt = 0,ans = 0,pt = 0;
        for(int i=1; i<=n; i++) {
            if(v[i] == 1) {
                ans++;
            } else if(v[i] == 2) {
                cnt++;
            }else{
                rt = i;
                pt++;
            }
        }
        if(pt>=2){
            printf("NO\n");
            continue;
        }
        if(v[1]!=1) {
            if(v[1] == n-1) {
                printf("YES\n");
            } else {
                printf("NO\n");
            }
        }else{
            if(cnt + ans +1 >= n && v[pf] <=2){
                int flag = 0;
                if(pt>0){
                    for(int i=1;i<=n;i++){
                        if(map[rt][i] == 1 && v[i]!=1){
                            flag++;
                        }
                    }
                }
                if(flag <= 1){
                    printf("YES\n");
                }else{
                    printf("NO\n");
                }

            }else{
                if(v[pf] == n-1){
                    printf("YES\n");
                    continue;
                }
                printf("NO\n");
            }
        }

    }
    return 0;
}