学步园

传送门：【HDU】1595 find the longest of the shortest

题目分析：首先求出一条最短路，记录下最短路上用到的边，枚举删除每一条边，求一次最短路，求完后恢复删除的边。重复这一过程直到枚举完所有的边为止。所有删除边后求得的最短路里最长的那条就是答案。

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 = 2000005 ;
const int MAXE = 2000005 ;
const int INF = 0x3f3f3f3f ;

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

struct Heap {
	int v ;
	int 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 x ) {
		int o = x , p = o , l = o << 1 , r = o << 1 | 1 ;
		while ( o > 1 && heap[o] < heap[o >> 1] ) {
			swap ( heap[o] , heap[o >> 1] ) ;
			o >>= 1 ;
		}
		o = x ;
		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 ) ;
		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 used[MAXN] ;
	int d[MAXN] ;
	int f[MAXN] ;
	bool vis[MAXN] ;
	
	void init () {
		cur = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , int c = 0 , bool flag = 0 ) {
		E[cur] = Edge ( v , c , flag , H[u] ) ;
		H[u] = cur ++ ;
	}
	
	void dijkstra ( int s , int t , bool first ) {
		CLR ( d , INF ) ;
		CLR ( vis , 0 ) ;
		q.clear () ;
		d[s] = 0 ;
		f[s] = -1 ;
		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 ;
				if ( !E[i].flag && d[v] > d[u] + E[i].c ) {
					if ( first ) {
						f[v] = u ;
						used[v] = i ;
					}
					d[v] = d[u] + E[i].c ;
					q.push ( d[v] , v ) ;
				}
			}
		}
	}
} dij ;

int n , m ;

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 ;
	dij.init () ;
	while ( m -- ) {
		scanf ( u ) , scanf ( v ) , scanf ( c ) ;
		dij.addedge ( u , v , c ) ;
		dij.addedge ( v , u , c ) ;
	}
	dij.dijkstra ( n , 1 , 1 ) ;
	int ans = dij.d[1] ;
	for ( int v = 1 ; ~v ; v = dij.f[v] ) {
		dij.E[dij.used[v]].flag = 1 ;
		dij.dijkstra ( n , 1 , 0 ) ;
		dij.E[dij.used[v]].flag = 0 ;
		ans = max ( ans , dij.d[1] ) ;
	}
	printf ( "%d\n" , ans ) ;
}

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