Codeforces 653E：Bear and Forgotten Tree 2

E. Bear and Forgotten Tree 2

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

A tree is a connected undirected graph consisting of n vertices and n  -  1 edges.
Vertices are numbered 1 through n.

Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.

You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm).
Limak remembers that for each i there was no edge between ai andbi.
He also remembers that vertex 1 was incident to exactly k edges
(its degree was equal to k).

Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.

Input

The first line of the input contains three integers n, m and k (

) —
the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.

The i-th of next m lines
contains two distinct integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) —
the i-th pair that is forbidden. It's guaranteed that each pair of
vertices will appear at most once in the input.

Output

Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible"
(without quotes).

Examples

input

5 4 2
1 2
2 3
4 2
4 1

output

possible

input

6 5 3
1 2
1 3
1 4
1 5
1 6

output

impossible

Note

In the first sample, there are n = 5 vertices. The degree of vertex 1 should
be k = 2. All conditions are satisfied for a tree with edges1 - 5, 5 - 2, 1 - 3 and 3 - 4.

In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6.
Hence, vertex 1couldn't be connected to any other vertex and it implies that there is no suitable tree.

题意是给出了n个节点，m个不相连的边，节点1的度确定为k。问给出的关系能否形成一棵树。

做法很亮。。。但后来想一想就是判断联通，联通的点就从set中去掉，然后去看不得不与1相连的点的个数，是否大于k。大于k就不行，否则ok。

代码：
#pragma warning(disable:4996)
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
using namespace std;
typedef long long ll;

#define INF 0x33ffffff

const ll mod = 1000000007;
const int maxn = 505;
const double PI = acos(-1.0);

int n, m, k;
set<int>remain;
set< pair<int, int> > forbidden;

void dfs(int x)
{
vector<int>m;
for (auto i : remain)
{
int mx = max(x, i);
int mi = min(x, i);
if (forbidden.find(make_pair(mi, mx)) == forbidden.end())
{
m.push_back(i);
}
}
for (auto i : m)
{
remain.erase(i);
}
for (auto i : m)
{
dfs(i);
}
}

void solve()
{
int i, j, u, v, com;
scanf("%d%d%d", &n, &m, &k);
for (i = 1, j = n - 1; i <= m; i++)
{
scanf("%d%d", &u, &v);
if (u > v)swap(u, v);
if (u == 1)j--;
forbidden.insert(make_pair(u, v));
}
if (j < k)
{
puts("impossible");
return;
}
for (i = 2; i <= n; i++)
{
remain.insert(i);
}
com = 0;
for (i = 2; i <= n; i++)
{
if (forbidden.find(make_pair(1, i)) == forbidden.end() && remain.find(i) != remain.end())
{
com++;
dfs(i);
remain.erase(i);
}
}
if (com > k || !remain.empty())
{
puts("impossible");
return;
}
puts("possible");
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("i.txt", "r", stdin);
freopen("o.txt", "w", stdout);
#endif

solve();

//system("pause");
return 0;
}