学步园

传送门：【HDU】1317 XYZZY

题目分析：首先我们可以用spfa判最长路上是否有正权环，但是有正权环却不等价于能到达终点。这是我们还需要判断是否能从正权环中走到终点，这个可以用传递闭包搞定。如果没有正权环就看是否能从起点到终点就好了。

代码如下：

#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 ) ; i >= ( b ) ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 105 ;
const int MAXE = 23333 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v ;
	//int c ;
	Edge* next ;
} E[MAXE] , *H[MAXN] , *cur ;

int d[MAXN] ;
int Q[MAXN] , head , tail ;
bool inq[MAXN] ;
int in[MAXN] ;
int n ;
int val[MAXN] ;
bool G[MAXN][MAXN] ;

void init () {
	cur = E ;
	CLR ( H , 0 ) ;
}

void addedge ( int u , int v , int c = 0 ) {
	cur -> v = v ;
	//cur -> c = c ;
	cur -> next = H[u] ;
	H[u] = cur ++ ;
}

bool spfa () {
	CLR ( in , 0 ) ;
	CLR ( inq , 0 ) ;
	CLR ( d , -INF ) ;
	head = tail = 0 ;
	Q[tail ++] = 1 ;
	d[1] = 100 ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		if ( head == MAXN ) head = 0 ;
		inq[u] = 0 ;
		for ( Edge* e = H[u] ; e ; e = e -> next ) {
			int v = e -> v ;
			if ( d[v] < d[u] + val[v] && d[u] + val[v] > 0 ) {
				d[v] = d[u] + val[v] ;
				if ( !inq[v] ) {
					if ( ++ in[v] > n ) return G[v][n] ;
					inq[v] = 1 ;
					Q[tail ++] = v ;
					if ( tail == MAXN ) tail = 0 ;
				}
			}
		}
	}
	return d[n] > 0 ;
}

void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void solve () {
	int m , v ;
	init () ;
	CLR ( G , 0 ) ;
	FOR ( i , 1 , n ) {
		scanf ( "%d" , &val[i] ) ;
		scanf ( "%d" , &m ) ;
		while ( m -- ) {
			scanf ( "%d" , &v ) ;
			addedge ( i , v ) ;
			G[i][v] = 1 ;
		}
	}
	FOR ( k , 1 , n ) FOR ( i , 1 , n ) FOR ( j , 1 , n ) G[i][j] |= G[i][k] & G[k][j] ;
	if ( spfa () ) printf ( "winnable\n" ) ;
	else printf ( "hopeless\n" ) ;
}

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