本文主要是介绍【HDU】1317 XYZZY spfa判负环+floyd求传递闭包，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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 ;
}这篇关于【HDU】1317 XYZZY spfa判负环+floyd求传递闭包的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
