【HDU】5044 Tree 树链剖分
传送门：【HDU】5044 Tree
题目分析：如果不看卡常数这回事的话。。这就是一个很裸的树链剖分。。然后就可以用线段树维护。
但是！这样对于本题是一定会超时的！因为出题人特意想卡。。。
于是我换成树状数组+输入优化卡过。。。
但这题还有更好的方法！我们可以在树链剖分上用标记法，每次对连续区间的位置L标记+v，位置R+1标记-v，最后扫一遍结果就出来了。。
O(nlog^2n)代码如下：
#include
#include
#include
#include
using namespace std ;
typedef long long LL ;
#pragma comment(linker, "/STACK:16777216")
#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 )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , n
const int MAXN = 100005 ;
const int MAXE = 200005 ;
struct Edge {
int v ;
Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;
struct Seg {
int u , v ;
Seg () {}
Seg ( int u , int v ) : u ( u ) , v ( v ) {}
} seg[MAXN] ;
int siz[MAXN] ;
int top[MAXN] ;
int pre[MAXN] ;
int pos[MAXN] ;
int dep[MAXN] ;
int son[MAXN] ;
int tree_idx ;
int ans1[MAXN] , ans2[MAXN] ;
int idx[MAXN] ;
int seg_idx[MAXN] ;
LL T[2][MAXN] ;
LL addv[2][MAXN << 2] ;
int n , q ;
void clear () {
edge = E ;
tree_idx = 0 ;
pre[1] = 0 ;
siz[0] = 0 ;
clr ( H , 0 ) ;
clr ( T , 0 ) ;
}
void addedge ( int u , int v ) {
edge -> v = v ;
edge -> next = H[u] ;
H[u] = edge ++ ;
}
void dfs ( int u ) {
siz[u] = 1 ;
son[u] = 0 ;
travel ( e , H , u ) {
int v = e -> v ;
if ( v != pre[u] ) {
pre[v] = u ;
dep[v] = dep[u] + 1 ;
dfs ( v ) ;
siz[u] += siz[v] ;
if ( siz[v] > siz[son[u]] ) son[u] = v ;
}
}
}
void rewrite ( int u , int top_element ) {
top[u] = top_element ;
pos[u] = ++ tree_idx ;
idx[tree_idx] = u ;
if ( son[u] ) rewrite ( son[u] , top_element ) ;
travel ( e , H , u ) {
int v = e -> v ;
if ( v != son[u] && v != pre[u]) rewrite ( v , v ) ;
}
}
void add ( int x , int o , int v ) {
while ( x ) {
T[o][x] += v ;
x -= x & -x ;
}
}
LL sum ( int x , int o , LL ans = 0 ) {
while ( x <= n ) {
ans += T[o][x] ;
x += x & -x ;
}
return ans ;
}
void pushdown ( int x , int o ) {
if ( addv[x][o] ) {
addv[x][ls] += addv[x][o] ;
addv[x][rs] += addv[x][o] ;
addv[x][o] = 0 ;
}
}
void sub_update ( int L , int R , int x , int v , int o , int l , int r ) {
if ( L <= l && r <= R ) {
addv[x][o] += v ;
return ;
}
int m = mid ;
pushdown ( x , o ) ;
if ( L <= m ) sub_update ( L , R , x , v , lson ) ;
if ( m < R ) sub_update ( L , R , x , v , rson ) ;
}
void update ( int x , int y , int o , int v ) {
while ( top[x] != top[y] ) {
if ( dep[top[x]] > dep[top[y]] ) {
add ( pos[x] , o , v ) ;
add ( pos[top[x]] - 1 , o , -v ) ;
//sub_update ( pos[top[x]] , pos[x] , o , v , root ) ;
x = pre[top[x]] ;
} else {
add ( pos[y] , o , v ) ;
add ( pos[top[y]] - 1 , o , -v ) ;
//sub_update ( pos[top[y]] , pos[y] , o , v , root ) ;
y = pre[top[y]] ;
}
}
if ( o && x == y ) return ;
if ( dep[x] > dep[y] ) {
add ( pos[x] , o , v ) ;
add ( pos[y] + o - 1 , o , -v ) ;
//sub_update ( pos[y] + o , pos[x] , o , v , root ) ;
} else {
add ( pos[y] , o , v ) ;
add ( pos[x] + o - 1 , o , -v ) ;
//sub_update ( pos[x] + o , pos[y] , o , v , root ) ;
}
}
void down ( int o , int l , int r ) {
if ( l == r ) {
ans1[idx[l]] = addv[0][o] ;
ans2[seg_idx[l]] = addv[1][o] ;
return ;
}
int m = mid ;
pushdown ( 0 , o ) ;
pushdown ( 1 , o ) ;
down ( lson ) ;
down ( rson ) ;
}
void scanf ( int& x , char c = 0 , int flag = 0 ) {
while ( ( c = getchar () ) != '-' && ( c < '0' || c > '9' ) ) ;
if ( c == '-' ) flag = 1 , x = 0 ;
else x = c - '0' ;
while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
if ( flag ) x = -x ;
}
void solve () {
char buf[10] ;
int u , v , w ;
clear () ;
scanf ( "%d%d" , &n , &q ) ;
rep ( i , 1 , n ) {
scanf ( u ) , scanf ( v ) ;
seg[i] = Seg ( u , v ) ;
addedge ( u , v ) ;
addedge ( v , u ) ;
}
dfs ( 1 ) ;
rewrite ( 1 , 1 ) ;
rep ( i , 1 , n ) {
u = seg[i].u ;
v = seg[i].v ;
if ( dep[u] < dep[v] ) seg_idx[pos[v]] = i ;
else seg_idx[pos[u]] = i ;
}
while ( q -- ) {
scanf ( "%s" , buf ) ;
scanf ( u ) , scanf ( v ) , scanf ( w ) ;
if ( buf[3] == '1' ) update ( u , v , 0 , w ) ;
else update ( u , v , 1 , w ) ;
}
For ( i , 1 , n ) ans1[idx[i]] = sum ( i , 0 ) ;
For ( i , 2 , n ) ans2[seg_idx[i]] = sum ( i , 1 ) ;
//down ( root ) ;
For ( i , 1 , n ) printf ( "%d%s" , ans1[i] , i < n ? " " : "" ) ;
printf ( "\n" ) ;
rep ( i , 1 , n ) printf ( "%d%s" , ans2[i] , i < n - 1 ? " " : "" ) ;
printf ( "\n" ) ;
}
int main () {
int T , cas = 0 ;
scanf ( "%d", &T ) ;
while ( T -- ) {
printf ( "Case #%d:\n" , ++ cas ) ;
solve () ;
}
return 0 ;
} O(nlogn + n)代码如下：
#include
#include
#include
#include
using namespace std ;
typedef long long LL ;
#pragma comment(linker, "/STACK:16777216")
#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 )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , n
const int MAXN = 100005 ;
const int MAXE = 200005 ;
struct Edge {
int v ;
Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;
struct Seg {
int u , v ;
Seg () {}
Seg ( int u , int v ) : u ( u ) , v ( v ) {}
} seg[MAXN] ;
int siz[MAXN] ;
int top[MAXN] ;
int pre[MAXN] ;
int pos[MAXN] ;
int dep[MAXN] ;
int son[MAXN] ;
int tree_idx ;
LL ans1[MAXN] , ans2[MAXN] ;
int idx[MAXN] ;
int seg_idx[MAXN] ;
LL a[2][MAXN] ;
int n , q ;
void clear () {
edge = E ;
tree_idx = 0 ;
pre[1] = 0 ;
siz[0] = 0 ;
clr ( H , 0 ) ;
clr ( a , 0 ) ;
}
void addedge ( int u , int v ) {
edge -> v = v ;
edge -> next = H[u] ;
H[u] = edge ++ ;
}
void dfs ( int u ) {
siz[u] = 1 ;
son[u] = 0 ;
travel ( e , H , u ) {
int v = e -> v ;
if ( v != pre[u] ) {
pre[v] = u ;
dep[v] = dep[u] + 1 ;
dfs ( v ) ;
siz[u] += siz[v] ;
if ( siz[v] > siz[son[u]] ) son[u] = v ;
}
}
}
void rewrite ( int u , int top_element ) {
top[u] = top_element ;
pos[u] = ++ tree_idx ;
idx[tree_idx] = u ;
if ( son[u] ) rewrite ( son[u] , top_element ) ;
travel ( e , H , u ) {
int v = e -> v ;
if ( v != son[u] && v != pre[u]) rewrite ( v , v ) ;
}
}
void update ( int x , int y , int o , int v ) {
while ( top[x] != top[y] ) {
if ( dep[top[x]] > dep[top[y]] ) {
a[o][pos[top[x]]] += v ;
a[o][pos[x] + 1] -= v ;
x = pre[top[x]] ;
} else {
a[o][pos[top[y]]] += v ;
a[o][pos[y] + 1] -= v ;
y = pre[top[y]] ;
}
}
if ( o && x == y ) return ;
if ( dep[x] > dep[y] ) {
a[o][pos[y] + o] += v ;
a[o][pos[x] + 1] -= v ;
} else {
a[o][pos[x] + o] += v ;
a[o][pos[y] + 1] -= v ;
}
}
void scanf ( int& x , char c = 0 , int flag = 0 ) {
while ( ( c = getchar () ) != '-' && ( c < '0' || c > '9' ) ) ;
if ( c == '-' ) flag = 1 , x = 0 ;
else x = c - '0' ;
while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
if ( flag ) x = -x ;
}
void solve () {
char buf[10] ;
int u , v , w ;
clear () ;
scanf ( "%d%d" , &n , &q ) ;
rep ( i , 1 , n ) {
scanf ( u ) , scanf ( v ) ;
seg[i] = Seg ( u , v ) ;
addedge ( u , v ) ;
addedge ( v , u ) ;
}
dfs ( 1 ) ;
rewrite ( 1 , 1 ) ;
rep ( i , 1 , n ) {
u = seg[i].u ;
v = seg[i].v ;
if ( dep[u] < dep[v] ) seg_idx[pos[v]] = i ;
else seg_idx[pos[u]] = i ;
}
while ( q -- ) {
scanf ( "%s" , buf ) ;
scanf ( u ) , scanf ( v ) , scanf ( w ) ;
if ( buf[3] == '1' ) update ( u , v , 0 , w ) ;
else update ( u , v , 1 , w ) ;
}
LL x = 0 , y = 0 ;
For ( i , 1 , n ) {
x += a[0][i] ;
ans1[idx[i]] = x ;
}
For ( i , 2 , n ) {
y += a[1][i] ;
ans2[seg_idx[i]] = y ;
}
For ( i , 1 , n ) printf ( "%I64d%s" , ans1[i] , i < n ? " " : "" ) ;
printf ( "\n" ) ;
rep ( i , 1 , n ) printf ( "%I64d%s" , ans2[i] , i < n - 1 ? " " : "" ) ;
printf ( "\n" ) ;
}
int main () {
int T , cas = 0 ;
scanf ( "%d", &T ) ;
while ( T -- ) {
printf ( "Case #%d:\n" , ++ cas ) ;
solve () ;
}
return 0 ;
}