本文主要是介绍【POJ】2449 Remmarguts' Date【k短路】，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！
题目链接：Remmarguts’ Date
k短路模板题，用这题验证了我可持久化左偏树的正确性。
#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std ;typedef long long LL ;
typedef pair < int , int > pii ;
typedef pair < LL , int > pli ;
typedef unsigned long long ULL ;#define clr( a , x ) memset ( a , x , sizeof a )
#define st first
#define ed secondconst int MAXN = 10005 ;
const int BLOCK = 22 ;
const LL INF = 1e18 ;namespace Leftist_Tree {struct Node {int l , r , x , h ;LL val ;} T[MAXN * 200] ;int Root[MAXN] ;int node_num ;int newnode ( const Node& o ) {T[node_num] = o ;return node_num ++ ;}void init () {node_num = 1 ;T[0].l = T[0].r = T[0].x = T[0].h = 0 ;T[0].val = INF ;}int merge ( int x , int y ) {if ( !x ) return y ;if ( T[x].val > T[y].val ) swap ( x , y ) ;int o = newnode ( T[x] ) ;T[o].r = merge ( T[o].r , y ) ;if ( T[T[o].l].h < T[T[o].r].h ) swap ( T[o].l , T[o].r ) ;T[o].h = T[T[o].r].h + 1 ;return o ;}void insert ( int& x , LL val , int v ) {int o = newnode ( T[0] ) ;T[o].val = val , T[o].x = v ;x = merge ( x , o ) ;}void show ( int o ) {printf ( "%d %lld %lld %lld\n" , o , T[o].val , T[T[o].l].val , T[T[o].r].val ) ;if ( T[o].l ) show ( T[o].l ) ;if ( T[o].r ) show ( T[o].r ) ;}
}using namespace Leftist_Tree ;
vector < pii > G[MAXN] , E[MAXN] ;
int vis[MAXN] ;
int in[MAXN] , p[MAXN] ;
LL d[MAXN] ;
int s , t ;
int n , m , k ;void addedge ( int u , int v , int c ) {G[u].push_back ( pii ( v , c ) ) ;E[v].push_back ( pii ( u , c ) ) ;
}void dij () {priority_queue < pli > q ;d[t] = 0 ;q.push ( pli ( 0 , t ) ) ;while ( !q.empty () ) {int u = q.top ().ed ;q.pop () ;if ( vis[u] ) continue ;vis[u] = 1 ;for ( int i = 0 ; i < E[u].size () ; ++ i ) {int v = E[u][i].st ;if ( d[v] > d[u] + E[u][i].ed ) {p[v] = u ;d[v] = d[u] + E[u][i].ed ;q.push ( pli ( -d[v] , v ) ) ;}}}
}void dfs ( int u ) {if ( vis[u] ) return ;vis[u] = 1 ;if ( p[u] ) Root[u] = Root[p[u]] ;int flag = 1 ;for ( int i = 0 ; i < G[u].size () ; ++ i ) {int v = G[u][i].st ;if ( d[v] == INF ) continue ;if ( p[u] == v && d[u] == G[u][i].ed + d[v] && flag ) {flag = 0 ;continue ;}LL val = d[v] - d[u] + G[u][i].ed ;insert ( Root[u] , val , v ) ;}for ( int i = 0 ; i < E[u].size () ; ++ i ) {if ( p[E[u][i].st] == u ) dfs ( E[u][i].st ) ;}
}void solve () {for ( int i = 1 ; i <= n ; ++ i ) {G[i].clear () ;E[i].clear () ;d[i] = INF ;vis[i] = 0 ;p[i] = 0 ;}for ( int i = 0 ; i < m ; ++ i ) {int u , v , c ;scanf ( "%d%d%d" , &u , &v , &c ) ;addedge ( u , v , c ) ;}scanf ( "%d%d%d" , &s , &t , &k ) ;dij () ;if ( d[s] == INF ) {printf ( "-1\n" ) ;return ;}if ( s != t ) -- k ;if ( !k ) {printf ( "%lld\n" , d[s] ) ;return ;}for ( int i = 1 ; i <= n ; ++ i ) {vis[i] = 0 ;}init () ;Root[t] = 0 ;dfs ( t ) ;priority_queue < pli , vector < pli > , greater < pli > > q ;if ( Root[s] ) q.push ( pli ( d[s] + T[Root[s]].val , Root[s] ) ) ;while ( k -- ) {if ( q.empty () ) {printf ( "-1\n" ) ;return ;}pli u = q.top () ;q.pop () ;if ( !k ) {printf ( "%lld\n" , u.st ) ;return ;}int x = T[u.ed].l , y = T[u.ed].r , v = T[u.ed].x ;if ( Root[v] ) q.push ( pli ( u.st + T[Root[v]].val , Root[v] ) ) ;if ( x ) q.push ( pli ( u.st + T[x].val - T[u.ed].val , x ) ) ;if ( y ) q.push ( pli ( u.st + T[y].val - T[u.ed].val , y ) ) ;}
}int main () {while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;return 0 ;
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