【POJ】1797 Heavy Transportation 二分+最短路_poj1797 heavy transportation 二分-优快云博客

本文链接：https://blog.youkuaiyun.com/u013368721/article/details/38819969

本文介绍了一个名为Heavy Transportation的算法竞赛题目。通过二分查找边权并利用最短路径算法求解，实现对特定条件下的路径搜索。代码示例中详细展示了自定义优先级队列、最短路径算法的具体实现细节。

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题目分析：二分边权然后做最短路。

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 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 = 1005 ;
const int MAXH = 1000005 ;
const int MAXE = 1000005 ;
const int INF = 0x3f3f3f3f ;

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

struct Heap {
	int v , idx ;
	Heap () {}
	Heap ( value_t v , int idx ) : v ( v ) , idx ( idx ) {}
	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 ) {
		h[point] = Heap ( v , idx ) ;
		maintain ( point ++ ) ;
	}
	void pop () {
		h[1] = h[-- point] ;
		maintain ( 1 ) ;
	}
	bool empty () {
		return point == 1 ;
	}
	int front () {
		return h[1].idx ;
	}
} ;

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

int n , m ;

void solve () {
	int u , v , c ;
	int l = INF , r = 0 ;
	G.clear () ;
	scanf ( "%d%d" , &n , &m ) ;
	while ( m -- ) {
		scanf ( "%d%d%d" , &u , &v , &c ) ;
		G.addedge ( u , v , c ) ;
		G.addedge ( v , u , c ) ;
		if ( l > c ) l = c ;
		if ( c > r ) r = c ;
	}
	while ( l < r ) {
		int mid = ( l + r + 1 ) >> 1 ;
		if ( G.dijkstra ( 1 , n , mid ) ) l = mid ;
		else r = mid - 1 ;
	}
	printf ( "%d\n\n" , l ) ;
}

int main () {
	int T , cas = 0 ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		printf ( "Scenario #%d:\n" , ++ cas ) ;
		solve () ;
	}
	return 0 ;
}