1069. Prufer Code
Time limit: 0.25 second
Memory limit: 8 MB
Memory limit: 8 MB
A tree (i.e. a connected graph without cycles) with vertices is given (N ≥ 2). Vertices of the tree are numbered by the integers 1,…,N. A Prufer code for the tree is built as follows: a leaf (a vertex that is incident to the only edge) with a minimal number is taken. Then this vertex and the incident edge are removed from the graph, and the number of the vertex that was adjacent to the leaf is written down. In the obtained graph once again a leaf with a minimal number is taken, removed and this procedure is repeated until the only vertex is left. It is clear that the only vertex left is the vertex with the number N. The written down set of integers (N−1 numbers, each in a range from 1 to N) is called a Prufer code of the graph.
Your task is, given a Prufer code, to reconstruct a tree, i.e. to find out the adjacency lists for every vertex in the graph.
You may assume that 2 ≤ N ≤ 7500
Input
A set of numbers corresponding to a Prufer code of some tree. The numbers are separated with a spaces and/or line breaks.
Output
Adjacency lists for each vertex. Format: a vertex number, colon, numbers of adjacent vertices separated with a space. The vertices inside lists and lists itself should be sorted by vertex number in an ascending order (look at sample output).
Sample
| input | output |
|---|---|
2 1 6 2 6 |
1: 4 6 |
Problem Author: Magaz Asanov
Problem Source: Ural State Univerisity Personal Contest Online February'2001 Students Session
Problem Source: Ural State Univerisity Personal Contest Online February'2001 Students Session
Difficulty: 522
题意：给出prufer code，要求求出原树。
分析：知道prufercode之后就很简单了。
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
int n, arr[N];
int cnt[N];
priority_queue<int> que;
vector<int> ans[N]; inline void Input()
{
int x;
while(cin >> x)
{
arr[n++] = x;
cnt[x]++;
}
n++;
} inline void Solve()
{
for(int i = ; i < n; i++)
if(!cnt[i + ]) que.push(-(i + ));
for(int step = ; step <= n - ; step++)
{
int x = -que.top(), now = arr[step - ];
que.pop();
if(!(--cnt[now])) que.push(-now);
ans[now].pub(x);
ans[x].pub(now);
} for(int i = ; i <= n; i++)
{
printf("%d:", i);
sort(ans[i].begin(), ans[i].end());
int length = sz(ans[i]);
for(int j = ; j < length; j++)
printf(" %d", ans[i][j]);
printf("\n");
}
} int main()
{
freopen("a.in", "r", stdin);
Input();
Solve();
return ;
}