【HDU】1535 Invitation Cards 最短路

传送门：【HDU】1535 Invitation Cards

题目分析：题目真难读。。。。。。其实题目的意思就是让求从编号为1的点到其他所有点的最短路距离之和加上其他所有点到编号为1的点的最短路距离之和。只要在原图以及反向图分别跑一次最短路就好了。。

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 )const int MAXN = 1000005 ;
const int MAXH = 2000005 ;
const int MAXE = 2000005 ;
const int INF = 0x3f3f3f3f ;struct Edge {int v , c ;Edge* next ;
} ;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] , *H[MAXN] , *cur ;Edge rE[MAXE] , *rH[MAXN] , *rcur ;int d[MAXN] ;bool vis[MAXN] ;int Q[MAXN] , head , tail ;void init () {cur = E ;rcur = rE ;CLR ( H , 0 ) ;CLR ( rH , 0 ) ;}void addedge ( int u , int v , int c ) {cur -> v = v ;cur -> c = c ;cur -> next = H[u] ;H[u] = cur ++ ;rcur -> v = u ;rcur -> c = c ;rcur -> next = rH[v] ;rH[v] = rcur ++ ;}void dijkstra ( int s , Edge* 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 ;travel ( e , H , u ) {int v = e -> v , c = e -> c ;if ( d[v] > d[u] + c ) {d[v] = d[u] + c ;q.push ( d[v] , v ) ;}}}}void spfa ( int s , Edge* H[] ) {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 ;travel ( e , H , u ) {int v = e -> v , c = e -> 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 ;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 ans = 0 ;G.init () ;scanf ( n ) , scanf ( m ) ;while ( m -- ) {scanf ( u ) , scanf ( v ) , scanf ( c ) ;G.addedge ( u , v , c ) ;}G.dijkstra ( 1 , G.H ) ;FOR ( i , 1 , n ) ans += G.d[i] ;G.dijkstra ( 1 , G.rH ) ;FOR ( i , 1 , n ) ans += G.d[i] ;printf ( "%d\n" , ans ) ;
}int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;
}

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