【HDU】2363 Cycling 最短路

传送门：【HDU】2363 Cycling

题目分析：我写复杂了啊啊啊啊！！！看别人飞一般的速度，我的却跑的和乌龟似的。。。不禁嫩牛满面。。。

好了说一下我的思路吧，首先找到最大的高度差，作为二分的上限，然后二分高度差，再枚举每个点作为高度的下限，以这个点加上高度差作为高度上限跑最短路，如果用在这个高度范围内的点能存在最短路（枚举完所有点以后），那么就调整上界，同时顺便记录最短路长度，否则调整下界。

就这么简单。。。所以跑的慢 T U T

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 = 105 ;const int MAXH = 10005 ;const int MAXE = 10005 ;const int INF = 0x3f3f3f3f ;struct Edge {int v , c , n ;Edge () {}Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}} ;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 Q[MAXN] , head , tail ;void init () {cur = 0 ;CLR ( H , -1 ) ;}void addedge ( int u , int v , int c ) {E[cur] = Edge ( v , c , H[u] ) ;H[u] = cur ++ ;}int dijkstra ( int s , int t , int high[] , int l , int r ) {if ( high[s] < l || high[s] > r ) return INF ;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 ( high[v] < l || high[v] > r ) continue ;if ( d[v] > d[u] + c ) {d[v] = d[u] + c ;q.push ( d[v] , v ) ;}}}return d[t] ;}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 high[MAXN] ;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 ;int maxv = 0 , minv = INF ;int ans = 0 ;G.init () ;scanf ( "%d%d" , &n , &m ) ;FOR ( i , 1 , n ) {scanf ( high[i] ) ;if ( high[i] > maxv ) maxv = high[i] ;if ( high[i] < minv ) minv = high[i] ;}while ( m -- ) {scanf ( u ) , scanf ( v ) , scanf ( c ) ;G.addedge ( u , v , c ) ;G.addedge ( v , u , c ) ;}int l = 0 , r = maxv - minv + 1 ;while ( l < r ) {int mid = ( l + r ) >> 1 ;int tmp = INF ;FOR ( i , 1 , n ) {int low_limit = high[i] , high_limit = high[i] + mid ;tmp = min ( tmp , G.dijkstra ( 1 , n , high , low_limit , high_limit ) ) ;}if ( tmp != INF ) r = mid , ans = tmp ;else l = mid + 1 ;}printf ( "%d %d\n" , l , ans ) ;}int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;}