CF 141 div2 D (2-SAT)

E. The Road to Berland is Paved With Good Intentionstime limit per test

2 seconds

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

Berland hasNCities, some of them are connected by bidirectional roads. For each road we know whether it is asphalted or not.

The King of Berland Valera II wants to asphalt all roads of Berland, for that he gathered a group of workers. every day Valera chooses exactly one city and orders the crew to asphalt all roads that come from the city. the valiant crew fulfilled the King's order
In a day, then workers went home.

Unfortunately, not everything is as great as Valera II wowould like. the main part of the group were gastarbeiters-illegal immigrants who are enthusiastic but not exactly good at understanding orders in Berlandian. therefore, having stored ed orders to asphalt
The roads coming from some of the city, the group asphalted all non-asphalted roads coming from the city, and vice versa, took the asphalt from the roads that had it.

Upon learning of this progress, Valera II was very upset, but since it was too late to change anything, he asked you to make a program that determines whether you can in some way asphalt Berlandian roads in at mostNDays.
Help the king.

Input

The first line contains two space-separated integersN, Bytes,M-
The number of cities and roads in Berland, correspondingly. NextMLines contain the descriptions of roads in Berland:I-Th
Line contains three space-separated integersAI, Bytes,BI, Bytes,CI(1 digit ≤ DigitAI, Bytes,BILimit ≤ limitN;AI  =BI; 0 bytes ≤ bytesCILimit ≤ limit 1 ).
The first two integers (AI, Bytes,BI) Are
Indexes of the cities that are connected byI-Th road, the third integer (CI) Equals
1, if the road was initially asphalted, and 0 otherwise.

Consider the cities in Berland indexed from 1N, And the roads indexed from 1M.
It is guaranteed that between two Berlandian cities there is not more than one road.

Output

In the first line print a single integerX(0 bytes ≤ bytesXLimit ≤ limitN)-
The number of days needed to asphalt all roads. In the second line printXSpace-separated integers-the indexes of the cities to send the workers.
Print the cities in the order, in which Valera send the workers to asphalt roads. If there are multiple solutions, print any of them.

If there's no way to asphalt all roads, print "Impossible" (without the quotes ).

Sample test (s) input

4 41 2 12 4 04 3 13 2 0

Output

43 2 1 3

Input

3 31 2 02 3 03 1 0

Output

Impossible

First, each vertex can only be selected once at most, so the requirements for completion within n days are useless, and the order of city selection has no effect on the final result. Then, for an edge, if it is 1, either select the city adjacent to it, or do not select either. If it is 0, only one of the two cities can be selected. This is the 2SAT model. Apply the template directly. The Edge adding in the template meets the condition A or B. Here, we need A and B. just re-write an edge adding function.

#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<algorithm>#include<iostream>using namespace std;const int maxn = 100;struct TwoSAT {  int n;  vector<int> G[maxn*2];  bool mark[maxn*2];  int S[maxn*2], c;  bool dfs(int x) {    if (mark[x^1]) return false;    if (mark[x]) return true;    mark[x] = true;    S[c++] = x;    for (int i = 0; i < G[x].size(); i++)      if (!dfs(G[x][i])) return false;    return true;  }  void init(int n) {    this->n = n;    for (int i = 0; i < n*2; i++) G[i].clear();    memset(mark, 0, sizeof(mark));  }  // x = xval or y = yval  void add_clause(int x, int xval, int y, int yval) {    x = x * 2 + xval;    y = y * 2 + yval;    G[x^1].push_back(y);    G[y^1].push_back(x);  }  void new_add(int x,int xval,int y,int yval){    x = x*2 + xval;    y = y*2 + yval;    G[x].push_back(y);    G[y].push_back(x);  }  bool solve() {    for(int i = 0; i < n*2; i += 2)      if(!mark[i] && !mark[i+1]) {        c = 0;        if(!dfs(i)) {          while(c > 0) mark[S[--c]] = false;          if(!dfs(i+1)) return false;        }      }    return true;  }};TwoSAT solver;int main(){    int n,m;    while(cin >> n >> m){        solver.init(n);        for(int i = 0;i < m;i++){            int from,to,val;            cin >> from >> to >> val;            from--;to--;            if(val == 1){                solver.new_add(from,1,to,1);                solver.new_add(from,0,to,0);            }            else{                solver.new_add(from,1,to,0);                solver.new_add(from,0,to,1);            }        }        if(!solver.solve()){            cout << "Impossible\n";            continue;        }        int ans = 0;        for(int i = 1;i < 2*n;i+=2)            if(solver.mark[i] == true) ans++;        cout << ans << endl;        for(int i = 1;i < 2*n;i+=2)            if(solver.mark[i] == true) cout << i/2+1 << ' ';        cout << endl;    }    return 0;}