【HDU】2962 Trucking 二分+最短路

传送门：【HDU】2962 Trucking

题目分析：很直白的一道题，满足容量最大的情况下再求最短路，容量就二分吧，如果在该容量下存在最短路就调整下界，否则调整上界。

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 = 1005 ;
const int MAXH = 100005 ;
const int MAXE = 100005 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , c , n ;
	int h ;
	Edge () {}
	
	Edge ( int v , int c , int n , int h ) : v ( v ) , c ( c ) , n ( n ) , h ( h ) {}
} ;

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

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

	int front () {
		return heap[1].idx ;
	}
	
	Heap top () {
		return heap[1] ;
	}
} ;

struct Shortest_Path_Algorithm {
	priority_queue q ;
	Edge E[MAXE] ;
	int H[MAXN] , cur ;
	int d[MAXN] ;
	bool vis[MAXN] ;
	int used[MAXN] ;
	int f[MAXN] ;
	int Q[MAXN] , head , tail ;
	
	void init () {
		cur = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , int c , int h ) {
		E[cur] = Edge ( v , c , H[u] , h ) ;
		H[u] = cur ++ ;
	}
	
	bool dijkstra ( int s , int t , int h ) {
		q.clear () ;
		CLR ( d , INF ) ;
		CLR ( vis , 0 ) ;
		d[s] = 0 ;
		q.push ( d[s] , s ) ;
		while ( !q.empty () ) {
			int u = q.front () ;
			q.pop () ;
			if ( vis[u] ) continue ;
			vis[u] = 1 ;
			for ( int i = H[u] ; ~i ; i = E[i].n ) {
				int v = E[i].v , c = E[i].c ;
				if ( E[i].h >= h && d[v] > d[u] + c ) {
					d[v] = d[u] + c ;
					q.push ( d[v] , v ) ;
				}
			}
		}
		return d[t] != INF ;
	}
	
	void spfa ( int s , int t ) {
		head = tail = 0 ;
		CLR ( d , INF ) ;
		CLR ( vis , 0 ) ;
		d[s] = 0 ;
		Q[tail ++] = s ;
		while ( head != tail ) {
			int u = Q[head ++] ;
			if ( head == MAXN ) head = 0 ;
			vis[u] = 0 ;
			for ( int i = H[u] ; ~i ; i = E[i].n ) {
				int v = E[i].v , c = E[i].c ;
				if ( d[v] > d[u] + c ) {
					d[v] = d[u] + c ;
					if ( !vis[v] ) {
						vis[v] = 1 ;
						if ( d[v] < d[Q[head]] ) {
							if ( head == 0 ) head = MAXN ;
							Q[-- head] = v ;
						} else {
							Q[tail ++] = v ;
							if ( tail == MAXN ) tail = 0 ;
						}
					}
				}
			}
		}
	}
} G ;

int n , m ;
int s , t , limit ;

/*
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 u , v , c , h ;
	int ans = -1 ;
	G.init () ;
	while ( m -- ) {
		scanf ( "%d%d%d%d" , &u , &v , &h , &c ) ;
		if ( h == -1 ) h = INF ;
		G.addedge ( u , v , c , h ) ;
		G.addedge ( v , u , c , h ) ;
	}
	scanf ( "%d%d%d" , &s , &t , &limit ) ;
	int l = 0 , r = limit ;
	while ( l < r ) {
		int mid = ( l + r + 1 ) >> 1 ;
		if ( G.dijkstra ( s , t , mid ) ) ans = G.d[t] , l = mid ;
		else r = mid - 1 ;
	}
	if ( ~ans ) {
		printf ( "maximum height = %d\n" , l ) ;
		printf ( "length of shortest route = %d\n" , ans ) ;
	} else printf ( "cannot reach destination\n" ) ;
}

int main () {
	int cas = 0 ;
	while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) {
		if ( cas ) printf ( "\n" ) ;
		printf ( "Case %d:\n" , ++ cas ) ;
		solve () ;
	}
	return 0 ;
}