传送门:【HDU】3986 Harry Potter and the Final Battle
题目分析:先求一次最短路,同时记录在最短路上的顶点以及以该顶点为弧尾的最短路上的边。然后枚举删除每一条边,分别求一次最短路,其中最大的即答案。当然不可达输出-1。
测试发现堆优化的dij不如slf优化的spfa。。可能图太稀疏了吧。。。反正我觉得我写的挺搓的了。。。
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 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 , int closed ) { 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 ( i != closed && d[v] > d[u] + c ) { d[v] = d[u] + c ; q.push ( d[v] , v ) ; if ( closed == -1 ) { f[v] = u ; used[v] = i ; } } } } } void spfa ( int s , int t , int closed ) { 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 ( i != closed && 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 ; } } if ( closed == -1 ) { f[v] = u ; used[v] = i ; } } } } } } 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 ; G.init () ; scanf ( n ) ; scanf ( m ) ; while ( m -- ) { scanf ( u ) , scanf ( v ) , scanf ( c ) ; G.addedge ( u , v , c ) ; G.addedge ( v , u , c ) ; } G.spfa ( 1 , n , -1 ) ; if ( G.d[n] == INF ) { printf ( "-1\n" ) ; return ; } int ans = G.d[n] ; for ( int i = n ; i != 1 ; i = G.f[i] ) { G.spfa ( 1 , n , G.used[i] ) ; if ( G.d[n] > ans ) ans = G.d[n] ; if ( ans == INF ) break ; } printf ( "%d\n" , ans == INF ? -1 : ans ) ; } int main () { int T ; scanf ( "%d" , &T ) ; while ( T -- ) solve () ; return 0 ; }