【HDU】1142 A Walk Through the Forest 最短路+记忆化搜索

传送门：【HDU】1142 A Walk Through the Forest

题目分析：总感觉这题写过。。但是poj和hdu上都出奇的没有AC。本题并不是计算最短路的数量，这点要明晰。本题要求的是满足条件的路的数量。举个栗子：如果存在如下四条边，(1,3,2),(3,2,2),(1,4,3),(4,2,3)，满足条件的路不是1条而是两条：1->3->2，1->4->2。所以从终点求一次最短路以后再从终点寻找满足条件的路记忆化搜索，边界条件是dp[1] = 1。

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 = 100005 ;const int MAXE = 100005 ;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 used[MAXN] ;int f[MAXN] ;int Q[MAXN] , head , tail ;void init () {cur = 0 ;CLR ( H , -1 ) ;}void addedge ( int u , int v , int c = 0 ) {E[cur] = Edge ( v , c , H[u] ) ;H[u] = cur ++ ;}void dijkstra ( int s , int t ) {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 ( d[v] > d[u] + c ) {d[v] = d[u] + c ;q.push ( d[v] , v ) ;}}}}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 dp[MAXN] ;bool vis[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' ;}int dfs ( int u ) {if ( !vis[u] ) {vis[u] = 1 ;dp[u] = 0 ;for ( int i = G.H[u] ; ~i ; i = G.E[i].n )if ( G.d[G.E[i].v] > G.d[u] )dp[u] += dfs ( G.E[i].v ) ;}return dp[u] ;}void solve () {int u , v , c ;G.init () ;CLR ( vis , 0 ) ;while ( m -- ) {scanf ( "%d%d%d" , &u , &v , &c ) ;G.addedge ( u , v , c ) ;G.addedge ( v , u , c ) ;}G.spfa ( 2 , 1 ) ;vis[1] = 1 ;dp[1] = 1 ;printf ( "%d\n" , dfs ( 2 ) ) ;}int main () {while ( ~scanf ( "%d%d" , &n , &m ) && n ) solve () ;return 0 ;}