題目分析：01分數規劃+最小割。因為要求r = sigma(x*c/x*k)（x取0或1）最小，那麼不妨設g(r) = x*c - r*x*k。由01規劃的性質，我們可知：g(r)=0時是最優解，當g(r)<0時r可以繼續減小，調整二分上界，當g(r)>0時的r不符合要求，調整二分下界。我不知道網路流中出現負權邊會怎麼樣，不過還是最好不要有，那麼每條邊變成c-r以後，如果c-r<=0，那麼直接加到res裡，否則建到圖中跑最小割。跑完最小割以後res+=flow，如果res<=0，那麼調整上界，否則調整下界。最後要求的邊就是最後一次二分時的負權邊加上割邊，割邊可以dfs求出。

表示我的語文已經掛了。。。沒聽懂的可以看看最小割那啥論文，當然最好看看01分數規劃的相關資料。。。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 110 ;
const int MAXQ = 110 ;
const int MAXE = 1000 ;
const int INF = 0x3f3f3f3f ;
const double eps = 1e-8 ;

struct Edge {
	int v , n ;
	double c ;
	Edge () {}
	Edge ( int v , double c , int n ) : v ( v ) , c ( c ) , n ( n ) {}
} ;

struct Line {
	int u , v , c ;
	void input () {
		scanf ( "%d%d%d" , &u , &v , &c ) ;
	}
} ;

struct NetWork {
	Edge E[MAXE] ;
	int H[MAXN] , cntE ;
	int d[MAXN] , num[MAXN] , pre[MAXN] , cur[MAXN] ;
	int Q[MAXN] , head , tail ;
	int s , t , nv ;
	double flow ;
	int n , m ;
	
	Line L[MAXE] ;
	int ans[MAXE] ;
	bool vis[MAXN] ;
	
	void init () {
		cntE = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , double c ) {
		E[cntE] = Edge ( v , c , H[u] ) ;
		H[u] = cntE ++ ;
		E[cntE] = Edge ( u , c , H[v] ) ;
		H[v] = cntE ++ ;
	}
	
	void rev_bfs () {
		CLR ( d , -1 ) ;
		CLR ( num , 0 ) ;
		head = tail = 0 ;
		Q[tail ++] = t ;
		d[t] = 0 ;
		num[d[t]] = 1 ;
		while ( head != tail ) {
			int u = Q[head ++] ;
			for ( int i = H[u] ; ~i ; i = E[i].n ) {
				int v = E[i].v ;
				if ( ~d[v] )
					continue ;
				d[v] = d[u] + 1 ;
				num[d[v]] ++ ;
				Q[tail ++] = v ;
			}
		}
	}
	
	double ISAP () {
		CPY ( cur , H ) ;
		rev_bfs () ;
		flow = 0 ;
		int u = pre[s] = s ;
		while ( d[s] < nv ) {
			if ( u == t ) {
				double f = INF ;
				int pos ;
				for ( int i = s ; i != t ; i = E[cur[i]].v )
					if ( f > E[cur[i]].c ) {
						f = E[cur[i]].c ;
						pos = i ;
					}
				for ( int i = s ; i != t ; i = E[cur[i]].v ) {
					E[cur[i]].c -= f ;
					E[cur[i] ^ 1].c += f ;
				}
				u = pos ;
				flow += f ;
			}
			for ( int &i = cur[u] ; ~i ; i = E[i].n )
				if ( E[i].c && d[u] == d[E[i].v] + 1 )
					break ;
			if ( ~cur[u] ) {
				pre[E[cur[u]].v] = u ;
				u = E[cur[u]].v ;
			}
			else {
				if ( 0 == ( -- num[d[u]] ) )
					break ;
				int mmin = nv ;
				for ( int i = H[u] ; ~i ; i = E[i].n )
					if ( E[i].c && mmin > d[E[i].v] ) {
						mmin = d[E[i].v] ;
						cur[u] = i ;
					}
				d[u] = mmin + 1 ;
				num[d[u]] ++ ;
				u = pre[u] ;
			}
		}
		return flow ;
	}
	
	int sgn ( double x ) {
		return ( x > eps ) - ( x < -eps ) ;
	}
	
	int ok ( double mid ) {
		double res = 0 ;
		init () ;
		REP ( i , 0 , m ) {
			double c = L[i].c - mid ;
			if ( sgn ( c ) <= 0 )
				res += c ;
			else
				addedge ( L[i].u , L[i].v , c ) ;
		}
		res += ISAP () ;
		return sgn ( res ) <= 0 ;
	}
	
	void dfs ( int u ) {
		vis[u] = 1 ;
		for ( int i = H[u] ; ~i ; i = E[i].n )
			if ( sgn ( E[i].c ) && !vis[E[i].v] )
				dfs ( E[i].v ) ;
	}
	
	void solve () {
		s = 1 ;
		t = n ;
		nv = t + 1 ;
		REP ( i , 0 , m )
			L[i].input () ;
		double l = 0 , r = 1e9 , mid ;
		while ( r - l > eps ) {
			mid = ( r + l ) / 2 ;
			if ( ok ( mid ) )
				r = mid ;
			else
				l = mid ;
		}
		int cnt = 0 ;
		CLR ( vis , 0 ) ;
		dfs ( 1 ) ;
		REP ( i , 0 , m ) {
			if ( sgn ( L[i].c - r ) <= 0 )
				ans[cnt ++] = i ;
			else if ( vis[L[i].u] != vis[L[i].v] )
				ans[cnt ++] = i ;
		}
		printf ( "%d\n" , cnt ) ;
		REP ( i , 0 , cnt )
			printf ( "%d%c" , ans[i] + 1 , i < cnt - 1 ? ' ' : '\n' ) ;
	}
} x ;

int main () {
	while ( ~scanf ( "%d%d" , &x.n , &x.m ) )
		x.solve () ;
	return 0 ;
}