传送门：【HDU】2807 The Shortest Path
题目分析：题目很简单，矩阵计算出两个城市的连通性，建边，然后每次询问求最短路回答（或者floyd预处理）。
当然暴力的代价是惨痛的，用堆优化+dij+输入优化最多800ms。
然后很好奇前面的是怎么跑的这么快的，看了别人写的题解才发现，原来他们是用了hash的方法将二维化为一维了，虽然可能会错误，但在出题人不是故意去卡的情况下算是基本可以高效的过这道题了。不过我宁可速度慢一点，求稳。
这里我两份代码均用了dij，因为我感觉不是稠密图。
自己一开始的方法：堆优化+dij+输入优化 G++796ms
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 = 100 ;
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 ;
struct Matrix {
int mat[MAXN][MAXN] , n ;
Matrix () {}
Matrix ( int n ) : n ( n ) {
CLR ( mat , 0 ) ;
}
Matrix operator * ( const Matrix& a ) const {
Matrix res ( n ) ;
FOR ( i , 0 , n ) FOR ( j , 0 , n ) FOR ( k , 0 , n ) res.mat[i][j] += mat[i][k] * a.mat[k][j] ;
return res ;
}
bool operator == ( const Matrix& a ) const {
FOR ( i , 0 , n ) FOR ( j , 0 , n ) if ( mat[i][j] != a.mat[i][j] ) return 0 ;
return 1 ;
}
} a[MAXN] ;
int n , m , q ;
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 ;
G.init () ;
FOR ( i , 1 , n ) a[i].n = m ;
FOR ( i , 1 , n ) REP ( x , 0 , m ) REP ( y , 0 , m ) scanf ( a[i].mat[x][y] ) ;
FOR ( i , 1 , n ) FOR ( j , 1 , n ) {
if ( i == j ) continue ;
Matrix tmp = a[i] * a[j] ;
FOR ( k , 1 , n ) if ( i != k && j != k ) if ( tmp == a[k] ) G.addedge ( i , k , 1 ) ;
}
scanf ( q ) ;
while ( q -- ) {
scanf ( u ) , scanf ( v ) ;
G.dijkstra ( u , v ) ;
if ( G.d[v] == INF ) printf ( "Sorry\n" ) ;
else printf ( "%d\n" , G.d[v] ) ;
}
}
int main () {
while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) solve () ;
return 0 ;
}
后来用了hash优化的代码：G++31ms
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 = 100 ;
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 , q ;
struct Matrix2D {
int mat[MAXN][MAXN] ;
} a[MAXN] ;
struct Matrix1D {
int mat[MAXN] ;
bool operator == ( const Matrix1D& a ) const {
REP ( i , 0 , m ) if ( mat[i] != a.mat[i] ) return 0 ;
return 1 ;
}
} b[MAXN] ;
Matrix1D mul ( Matrix2D a , Matrix1D b ) {
Matrix1D res ;
REP ( i , 0 , m ) {
res.mat[i] = 0 ;
REP ( j , 0 , m ) res.mat[i] += a.mat[i][j] * b.mat[j] ;
}
return res ;
}
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 ;
G.init () ;
FOR ( i , 1 , n ) REP ( j , 0 , m ) {
b[i].mat[j] = 0 ;
REP ( k , 0 , m ) {
scanf ( a[i].mat[j][k] ) ;
b[i].mat[j] += a[i].mat[j][k] * k ;
}
}
FOR ( i , 1 , n ) FOR ( j , 1 , n ) {
if ( i == j ) continue ;
Matrix1D tmp = mul ( a[i] , b[j] ) ;
FOR ( k , 1 , n ) if ( i != k && j != k ) if ( tmp == b[k] ) G.addedge ( i , k , 1 ) ;
}
scanf ( q ) ;
while ( q -- ) {
scanf ( u ) , scanf ( v ) ;
G.spfa ( u , v ) ;
if ( G.d[v] == INF ) printf ( "Sorry\n" ) ;
else printf ( "%d\n" , G.d[v] ) ;
}
}
int main () {
while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) solve () ;
return 0 ;
}