【POJ】1724 ROADS 二维最短路

传送门：【POJ】1724 ROADS

题目分析：老老实实的用最短路写了，虽然dfs或者bfs更快。。

最短路就需要二维了，d[v][cost]表示从起点到v且花费为cost的最短路的长度，然后就是普通的最短路了。效率很低。

代码如下：

#include <cmath>
#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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define CLR( a , x ) memset ( a , x , sizeof a )

typedef int value_t ;

const int MAXN = 105 ;
const int MAXH = 1000005 ;
const int MAXE = 1000005 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , w ;
	value_t c ;
	Edge* next ;
} ;

struct Heap {
	int v , idx , w ;
	Heap () {}
	Heap ( value_t v , int idx , int w ) : v ( v ) , idx ( idx ) , w ( w ) {}
	bool operator < ( const Heap& a ) const {
		return v < a.v ;
	}
} ;

struct priority_queue {
	Heap h[MAXH] ;
	int point ;
	priority_queue () : point ( 1 ) {}
	void clear () {
		point = 1 ;
	}
	void maintain ( int o ) {
		int p = o , l = o << 1 , r = o << 1 | 1 ;
		while ( o > 1 && h[o] < h[o >> 1] ) {
			swap ( h[o] , h[o >> 1] ) ;
			o >>= 1 ;
		}
		o = p ;
		while ( 1 ) {
			if ( l < point && h[l] < h[p] ) p = l ;
			if ( r < point && h[r] < h[p] ) p = r ;
			if ( o == p ) break ;
			swap ( h[o] , h[p] ) ;
			o = p , l = o << 1 , r = o << 1 | 1 ;
		}
	}
	void push ( value_t v , int idx , int w ) {
		h[point] = Heap ( v , idx , w ) ;
		maintain ( point ++ ) ;
	}
	void pop () {
		h[1] = h[-- point] ;
		maintain ( 1 ) ;
	}
	bool empty () {
		return point == 1 ;
	}
	int front () {
		return h[1].idx ;
	}
	Heap top () {
		return h[1] ;
	}
} ;

struct Shortest_Path_Algorithm {
	priority_queue q ;
	Edge E[MAXE] ;
	Edge* H[MAXN] ;
	Edge* cur ;
	value_t d[MAXN][10005] ;
	bool vis[MAXN][10005] ;
	void clear () {
		cur = E ;
		CLR ( H , 0 ) ;
	}
	void addedge ( int u , int v , value_t c , int w ) {
		cur -> v = v ;
		cur -> c = c ;
		cur -> w = w ;
		cur -> next = H[u] ;
		H[u] = cur ++ ;
	}
	void dijkstra ( int s , int K ) {
		q.clear () ;
		CLR ( vis , 0 ) ;
		CLR ( d , INF ) ;
		d[s][0] = 0 ;
		q.push ( d[s][0] , s , 0 ) ;
		while ( !q.empty () ) {
			Heap tmp = q.top () ;
			int u = tmp.idx ;
			int w = tmp.w ;
			q.pop () ;
			if ( vis[u][w] ) continue ;
			vis[u][w] = 1 ;
			travel ( e , H , u ) {
				int v = e -> v , cost = w + e -> w ;
				if ( cost <= K && d[v][cost] > d[u][w] + e -> c ) {
					d[v][cost] = d[u][w] + e -> c ;
					q.push ( d[v][cost] , v , cost ) ;
				}
			}
		}
	}
} G ;

int K , n , m ;

void solve () {
	int u , v , c , w ;
	G.clear () ;
	while ( m -- ) {
		scanf ( "%d%d%d%d" , &u , &v , &c , &w ) ;
		G.addedge ( u , v , c , w ) ;
	}
	G.dijkstra ( 1 , K ) ;
	int ans = INF ;
	FOR ( i , 0 , K ) if ( ans > G.d[n][i] ) ans = G.d[n][i] ;
	printf ( "%d\n" , ans == INF ? -1 : ans ) ;
}

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