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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".
(2 (6 (7)) (3) (5 (1) (4)) (8)) (1 (2 (3)))
5 2 5 2 6 2 8 2 3
应该是挺难的一道题了 队友敲了代码 然后就是我们三个无限的找错了
找了最少两个小时的错
最后终于在比赛截至前4分钟 A了 有时间自己再做~
#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
using namespace std;
struct Node
{
int num;
Node* next;
Node(){next=NULL;num=-1;}
Node(int x){next=NULL;num=x;}
}node[55];
int du[55];
char map[100005];
stack<int>s;
vector<int>res;
int main()
{
while(gets(map)!=NULL)
{
memset(du,0,sizeof(du));
int len=strlen(map);
int sum=0;
for(int i=0;i<len;i++)
{
if(map[i]=='('&&sum)
{
s.push(sum);
sum=0;
}
if(map[i]>='0'&&map[i]<='9')
{
sum=sum*10+(map[i]-'0');
}
if(map[i]==')')
{
if(sum)
{
// printf("sum=%d,top=%d\n",sum,s.top());
du[sum]++;
du[s.top()]++;
Node* p=&node[sum];
// printf("sum(%d)->next=\n",sum);
while(p->next)
{
// printf("\tnum=%d\n",p->num);
p=p->next;
}
p->next=new Node(s.top());
p=&node[s.top()];
// printf("top(%d)->next=\n",s.top());
while(p->next)
{
p=p->next;
// printf("\tnum=%d\n",p->num);
}
p->next=new Node(sum);
sum=0;
}
else
{
sum=s.top();
s.pop();
if(!s.empty())
{
du[sum]++;
du[s.top()]++;
// printf(")sum=%d,top=%d\n",sum,s.top());
Node* p=&node[sum];
// printf(")sum(%d)->next=\n",sum);
while(p->next)
{
p=p->next;
// printf("\t)%d ",p->num);
// getchar();
}
p->next=new Node(s.top());
p=&node[s.top()];
// printf(")top(%d)->next=\n",s.top());
while(p->next)
{
p=p->next;
// printf("\t)%d ",p->num);
// getchar();
}
p->next=new Node(sum);
}
sum=0;
}
}
}
// Node* p=&node[5];
// printf("5:\n");
// while(p->next)
// {
// p=p->next;
// printf("\t)%d ",p->num);
// }
// for(int i=0;i<9;i++)
// printf("%d\n",du[i]);
while(1)
{
int ok=0;
for(int i=0;i<55;i++)
{
// printf("du[%d]=%d\n",i,du[i]);
if(du[i]==1)
{
// printf("度数为1的点为%d\n",i);
ok=1;
res.push_back(node[i].next->num);
// printf("要输出的点为%d\n",node[i].next->num);
du[i]--;
du[node[i].next->num]--;
Node* p=&node[node[i].next->num];
Node* q=&node[node[i].next->num];
while(1)
{
p=p->next;
if(p->num==i)
{
q->next=p->next;
break;
}
q=p;
}
break;
}
}
if(!ok)
break;
}
for(int i=0;i<res.size();i++)
{
if(i==0)
printf("%d",res[i]);
else
printf(" %d",res[i]);
}
printf("\n");
res.resize(0);
while(!s.empty())
s.pop();
memset(node,0,sizeof(node));
}
return 0;
}
nyoj1254 Code the Tree (第七届河南省程序设计大赛)
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原文地址：http://blog.csdn.net/su20145104009/article/details/51519901
