【HDU】1245 Saving James Bond 最短路

传送门：【HDU】1245 Saving James Bond

题目分析：按照题目要求建图就好了，注意精度问题！C++能AC，G++可能会WA。

代码如下：

#include <cmath>#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 double type_c ;const int MAXN = 105 ;const int MAXH = 20005 ;const int MAXE = 20005 ;const double INF = 1e9 ;struct Edge {int v ;type_c c ;Edge* next ;} ;struct Point {int x , y ;Point () {}Point ( int x , int y ) : x ( x ) , y ( y ) {}Point operator - ( const Point& a ) const {return Point ( x - a.x , y - a.y ) ;}} ;struct Heap {type_c v ;int idx ;Heap () {}Heap ( type_c 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 ( o < point ) {if ( l < point && h[l] < h[p] ) p = l ;if ( r < point && h[r] < h[p] ) p = r ;if ( p == o ) break ;swap ( h[o] , h[p] ) ;o = p , l = o << 1 , r = o << 1 | 1 ;}}void push ( type_c 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] , *H[MAXN] , *cur ;type_c d[MAXN] ;bool vis[MAXN] ;int Q[MAXN] , head , tail ;int step[MAXN] ;void init () {cur = E ;CLR ( H , 0 ) ;}void addedge ( int u , int v , type_c c ) {cur -> v = v ;cur -> c = c ;cur -> next = H[u] ;H[u] = cur ++ ;}void dijkstra ( int s ) {q.clear () ;CLR ( vis , 0 ) ;CLR ( step , 0 ) ;REP ( i , 0 , MAXN ) d[i] = 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 ( d[v] > d[u] + e -> c ) {d[v] = d[u] + e -> c ;step[v] = step[u] + 1 ;q.push ( d[v] , v ) ;} else if ( d[v] == d[u] + e -> c ) step[v] = min ( step[v] , step[u] + 1 ) ;}}}void spfa ( int s ) {head = tail = 0 ;REP ( i , 0 , MAXN ) d[i] = 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 ;if ( d[v] > d[u] + e -> c ) {d[v] = d[u] + e -> 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 , D ;Point p[MAXN] ;void scanf ( int& x , char c = 0 , bool flag = 0 ) {while ( ( c = getchar () ) != '-' && ( c < '0' || c > '9' ) ) ;if ( c == '-' ) flag = 1 , c = getchar () ;x = c - '0' ;while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;if ( flag ) x = -x ;}double dist ( Point a ) {return sqrt ( ( double ) a.x * a.x + a.y * a.y ) ;}void solve () {if ( D >= 42.5 ) {printf ( "42.50 1\n" ) ;return ;}G.init () ;FOR ( i , 1 , n ) scanf ( p[i].x ) , scanf ( p[i].y ) ;FOR ( i , 1 , n ) {double d = dist ( p[i] ) - 7.5 ;if ( d <= D ) G.addedge ( 0 , i , d ) ;if ( 50 - abs ( p[i].x ) <= D ) G.addedge ( i , n + 1 , 50 - abs ( p[i].x ) ) ;if ( 50 - abs ( p[i].y ) <= D ) G.addedge ( i , n + 1 , 50 - abs ( p[i].y ) ) ;}FOR ( i , 1 , n ) FOR ( j , i + 1 , n ) {double d = dist ( p[i] - p[j] ) ;if ( D >= d ) {G.addedge ( i , j , d ) ;G.addedge ( j , i , d ) ;}}G.dijkstra ( 0 ) ;if ( G.d[n + 1] == INF ) printf ( "can't be saved\n" ) ;else printf ( "%.2f %d\n" , G.d[n + 1] , G.step[n + 1] ) ;}int main () {while ( ~scanf ( "%d%d" , &n , &D ) ) solve () ;return 0 ;}