BZOJ1975 [Sdoi2010]魔法猪学院 K短路 Astar A* 贪心

大家都很强， 可与之共勉 。

题意：
  给定一张图，一共有k的能量，求最多可以从1→N走多少次（路径不相同）？（每走一次都会消耗路径长的能量）。

题解：

  很明显是找出每一次都走还能走得最短路，这个贪心策略一定是正确的。所以我们用A∗算法。
  手写堆的话一定要开够数组……
  BZOJ把空间改小了mmp，

/**************************************************************    Problem: 1975    User: Lazer2001    Language: C++    Result: Accepted    Time:1356 ms    Memory:34572 kb****************************************************************/# include <bits/stdc++.h>template < class T >  inline bool chkmin ( T& d, const T& x )  {  return ( d > x ) ? ( d = x ), 1 : 0 ;  }# define N 5010# define M 500010template < class T >class Heap  {    private :        T a [1500010] ; int len ;    public :        inline bool empty ( )  {            return len == 0 ;        }        inline void clear ( )  {            len = 0 ;        }           inline void push ( const T& ele )  {            a [++ len] = ele ;            std :: push_heap ( a + 1, a + 1 + len, std :: greater < T > ( ) ) ;        }        inline T top ( )  {            return a [1] ;        }        inline void pop ( )  {            std :: pop_heap ( a + 1, a + 1 + len, std :: greater < T > ( ) ) ;            -- len ;        }} ;Heap < std :: pair < double, int > > Q ;struct edge  {    int to ;    double w ;    edge* nxt ;} ;struct Graph  {    edge* head [N], g [M], *cur ;    Graph ( )  {        cur = g ;        memset ( head, 0, sizeof head ) ;    }    inline void add_edge ( int u, int v, double w )  {        *cur = ( edge )  { v, w, head [u] } ;  head [u] = cur ++ ;    }    inline edge*& operator [] ( const int& u )  {  return head [u] ;  }} G, Rev ;double dis [N] ;inline void Dijkstra ( int S, int T )  {    std :: fill ( dis, dis + 1 + S, 1e10 ) ;    Q.clear ( ) ;    Q.push ( std :: make_pair ( dis [S] = 0, S ) ) ;     while ( ! Q.empty ( ) )  {        int u = Q.top ( ).second ; Q.pop ( ) ;        for ( edge* it = Rev [u] ; it ; it = it -> nxt )  {            if ( chkmin ( dis [it -> to], dis [u] + it -> w ) )  {                Q.push ( std :: make_pair ( dis [it -> to], it -> to ) ) ;            }        }    }}double sum ;inline int Astar ( int S, int T )  {    if ( dis [S] == 1e10 )  return 0 ;    int cnt ( 0 ) ;    Q.clear ( ) ;    Q.push ( std :: make_pair ( dis [S], S ) ) ;    while ( ! Q.empty ( ) )  {        int u = Q.top ( ).second ; double d = Q.top ( ).first ;  Q.pop ( ) ;        if ( u == T )  {            if ( sum - d >= 0.0 )  {                sum -= d ;                ++ cnt ;            }  else  return cnt ;        }        for ( edge* it = G [u] ; it ; it = it -> nxt )  {            Q.push ( std :: make_pair ( d - dis [u] + dis [it -> to] + it -> w, it -> to ) ) ;        }    }    return cnt ;}int main ( )  {//  freopen ( "in.txt", "r", stdin ) ;    int n, m ;    scanf ( "%d%d%lf", & n, & m, & sum ) ;    while ( m -- )  {        static int u, v ; static double w ;        scanf ( "%d%d%lf", & u, & v, & w ) ;        G.add_edge ( u, v, w ) ;        Rev.add_edge ( v, u, w ) ;    }    Dijkstra ( n, 1 ) ;    printf ( "%d\n", Astar ( 1, n ) ) ;    return 0 ;}