A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code of such a tree is built as follows: the leaf (a vertex that is incident to only one edge) with the minimal number is taken. This leaf, together with its incident edge is removed from the graph, while the number of the vertex that was adjacent to the leaf is written down. In the new obtained tree, this procedure is repeated, until there is only one vertex left (which, by the way, always has number n). The written down sequence of n-1 numbers is called the Prufer code of the tree.  Your task is, given a tree, to compute its Prufer code. The tree is denoted by a word of the language specified by the following grammar:

T ::= "(" N S ")"

S ::= " " T S  | empty

N ::= number

That is, trees have parentheses around them, and a number denoting the identifier of the root vertex, followed by arbitrarily many (maybe none) subtrees separated by a single space character. As an example, take a look at the tree in the figure below which is denoted in the first line of the sample input. To generate further sample input, you may use your solution to Problem 2568.  Note that, according to the definition given above, the root of a tree may be a leaf as well. It is only for the ease of denotation that we designate some vertex to be the root. Usually, what we are dealing here with is called an "unrooted tree".

The input contains several test cases. Each test case specifies a tree as described above on one line of the input file. Input is terminated by EOF. You may assume that 1<=n<=50

For each test case generate a single line containing the Prufer code of the specified tree. Separate numbers by a single space. Do not print any spaces at the end of the line.

第七届河南省赛

题解：

是一个叫什么Prufer树的东西，这个树有一定规律，就是每次找树枝节点最小的那个数的父节点，去掉这个枝；队友提供了思路，想了想敲了敲，就ac了，由于数字可以读取多位，错了几次；

代码：

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<set>

#include<stack>

using namespace std;

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define SL(x) scanf("%lld",&x)

#define  PI(x) printf("%d",x)

#define  PL(x) printf("%lld",x)

#define P_ printf(" ")

const int INF=0x3f3f3f3f;

const double PI=acos(-1.0);

typedef long long LL;

const int MAXN=1010;

char s[2010];

int mp[MAXN][MAXN];

int usd[MAXN];

int ans[MAXN];

int vis[MAXN];

int main(){

	stack<int>S;

	int t;

	while(mem(s,0),gets(s)){

		int mx=0;

		mem(usd,0);

		int len=strlen(s),temp=0;

		for(int i=0;i<len;i++){

			if(s[i]=='(')temp=1;

			if(isdigit(s[i])){

				if(temp){

					t=0;

					while(isdigit(s[i]))t=t*10+s[i]-'0',i++;

					mx=max(mx,t);

					if(!S.empty())mp[t][S.top()]=mp[S.top()][t]=1,usd[S.top()]++,usd[t]++;

					S.push(t);

					temp=0;

				}

			}

			if(s[i]==')')if(!S.empty())S.pop();

		}

		int k=0;

		mem(vis,0);

		int num=mx;

		for(int i=1;i<=mx;i++)usd[i]--;

		while(true){

			temp=INF;

		for(int i=1;i<=mx;i++)

				if(!vis[i])if(!usd[i])temp=min(i,temp);

			if(temp==INF)break;

			num--;

			for(int i=1;i<=mx;i++){

				if(!vis[i]&&mp[temp][i]){

					ans[k++]=i;break;

				}

			}

			vis[temp]=1;

			for(int i=1;i<=mx;i++)

			if(!vis[i]&&mp[temp][i])mp[temp][i]=mp[i][temp]=0,usd[i]--;

		}

		for(int i=0;i<k;i++){

			if(i)printf(" ");

			PI(ans[i]);

		}puts("");

	}

	return 0;

}