【HDU】5221 Occupation【树链剖分】

传送门：【HDU】5221 Occupation

题目分析：

最直接的想法，用一棵树链剖分维护路径，一棵dfs序线段树维护子树。因为每次最多修改一个点，所以修改的时候我们暴力修改每个点就可以了。

my  code:

#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std ;typedef long long LL ;#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 root 1 , 1 , n#define mid ( ( l + r ) >> 1 )const int MAXN = 100005 ;const int MAXE = 200005 ;const int INF = 0x3f3f3f3f ;struct Edge {    int v , n ;    Edge () {}    Edge ( int v , int n ) : v ( v ) , n ( n ) {}} ;Edge E[MAXE] ;int H[MAXN] , cntE ;int tpos[2][MAXN << 2] ;int tin[2][MAXN] ;int top[MAXN] ;int siz[MAXN] ;int pre[MAXN] ;int dep[MAXN] ;int pos[MAXN] ;int son[MAXN] ;int val[MAXN] ;int dfs_clock ;int idx[MAXN] ;int ou[MAXN] ;bool c[MAXN] ;int tree_idx ;int n , q ;int ans ;void init () {    ans = 0 ;    cntE = 0 ;    tree_idx = 0 ;    dfs_clock = 0 ;    clr ( H , -1 ) ;}void addedge ( int u , int v ) {    E[cntE] = Edge ( v , H[u] ) ;    H[u] = cntE ++ ;}void dfs ( int u ) {    idx[u] = ++ dfs_clock ;    tin[1][dfs_clock] = u ;    siz[u] = 1 ;    son[u] = 0 ;    for ( int i = H[u] ; ~i ; i = E[i].n ) {        int v = E[i].v ;        if ( v == pre[u] ) continue ;        pre[v] = u ;        dep[v] = dep[u] + 1 ;        dfs ( v ) ;        siz[u] += siz[v] ;        if ( siz[son[u]] < siz[v] ) son[u] = v ;    }    ou[u] = dfs_clock ;}void rebuild ( int u , int top_element ) {    top[u] = top_element ;    pos[u] = ++ tree_idx ;    tin[0][tree_idx] = u ;    if ( son[u] ) rebuild ( son[u] , top_element ) ;    for ( int i = H[u] ; ~i ; i = E[i].n ) {        int v = E[i].v ;        if ( v != pre[u] && v != son[u] ) rebuild ( v , v ) ;    }}void build ( int o , int l , int r ) {    if ( l == r ) {        tpos[0][o] = tin[0][l] ;        tpos[1][o] = tin[1][l] ;        return ;    }    int m = mid ;    build ( lson ) ;    build ( rson ) ;    tpos[0][o] = max ( tpos[0][ls] , tpos[0][rs] ) ;    tpos[1][o] = max ( tpos[1][ls] , tpos[1][rs] ) ;}void update ( int x , int i , int v , int o , int l , int r ) {    if ( l == r ) {        tpos[i][o] = v ? tin[i][l] : 0 ;        return ;    }    int m = mid ;    if ( x <= m ) update ( x , i , v , lson ) ;    else update ( x , i , v , rson ) ;    tpos[i][o] = max ( tpos[i][ls] , tpos[i][rs] ) ;}int query ( int L , int R , int i , int o , int l , int r ) {    if ( L <= l && r <= R ) return tpos[i][o] ;    int m = mid ;    if ( R <= m ) return query ( L , R , i , lson ) ;    if ( m <  L ) return query ( L , R , i , rson ) ;    return max ( query ( L , R , i , lson ) , query ( L , R , i , rson ) ) ;}void Update1 ( int x , int y ) {    while ( top[x] != top[y] ) {        if ( dep[top[x]] < dep[top[y]] ) swap ( x , y ) ;        while ( 1 ) {            int t = query ( pos[top[x]] , pos[x] , 0 , root ) ;            if ( !t ) break ;            c[t] = 0 ;            update ( pos[t] , 0 , 0 , root ) ;            update ( idx[t] , 1 , 0 , root ) ;            ans += val[t] ;            //printf ( "%d\n" , t ) ;        }        x = pre[top[x]] ;    }    if ( dep[x] > dep[y] ) swap ( x , y ) ;    while ( 1 ) {        int t = query ( pos[x] , pos[y] , 0 , root ) ;        if ( !t ) break ;        c[t] = 0 ;        update ( pos[t] , 0 , 0 , root ) ;        update ( idx[t] , 1 , 0 , root ) ;        ans += val[t] ;    }}void Update2 ( int L , int R ) {    while ( 1 ) {        int t = query ( L , R , 1 , root ) ;        if ( !t ) break ;        c[t] = 0 ;        update ( pos[t] , 0 , 0 , root ) ;        update ( idx[t] , 1 , 0 , root ) ;        ans += val[t] ;    }}void solve () {    int u , v , op ;    init () ;    scanf ( "%d" , &n ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        scanf ( "%d" , &val[i] ) ;        c[i] = 1 ;    }    for ( int i = 1 ; i < n ; ++ i ) {        scanf ( "%d%d" , &u , &v ) ;        addedge ( u , v ) ;        addedge ( v , u ) ;    }    dfs ( 1 ) ;    rebuild ( 1 , 1 ) ;    build ( root ) ;    scanf ( "%d" , &q ) ;    for ( int i = 0 ; i < q ; ++ i ) {        scanf ( "%d" , &op ) ;        if ( op == 1 ) {            scanf ( "%d%d" , &u , &v ) ;            Update1 ( u , v ) ;        } else if ( op == 2 ) {            scanf ( "%d" , &u ) ;            if ( !c[u] ) {                c[u] = 1 ;                update ( pos[u] , 0 , 1 , root ) ;                update ( idx[u] , 1 , 1 , root ) ;                ans -= val[u] ;            }        } else {            scanf ( "%d" , &u ) ;            Update2 ( idx[u] , ou[u] ) ;        }        printf ( "%d\n" , ans ) ;    }}int main () {    int T ;    scanf ( "%d" , &T ) ;    for ( int i = 1 ; i <= T ; ++ i ) {        solve () ;    }    return 0 ;}

---

然而我们可以发现，树链剖分是有dfs序，然后我们就可以发现对于一个节点的子树，其子树构成的区间也是连续的，那么我们只需要一个树链剖分就够了。

my  code:

#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std ;typedef long long LL ;#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 root 1 , 1 , n#define mid ( ( l + r ) >> 1 )const int MAXN = 100005 ;const int MAXE = 200005 ;const int INF = 0x3f3f3f3f ;struct Edge {    int v , n ;    Edge () {}    Edge ( int v , int n ) : v ( v ) , n ( n ) {}} ;Edge E[MAXE] ;int H[MAXN] , cntE ;int tpos[MAXN << 2] ;int tree[MAXN] ;int top[MAXN] ;int siz[MAXN] ;int pre[MAXN] ;int dep[MAXN] ;int pos[MAXN] ;int son[MAXN] ;int val[MAXN] ;int idx[MAXN] ;int dfs_clock ;bool c[MAXN] ;int tree_idx ;int in[MAXN] ;int ou[MAXN] ;int n , q ;int ans ;void init () {    ans = 0 ;    cntE = 0 ;    tree_idx = 0 ;    dfs_clock = 0 ;    clr ( H , -1 ) ;}void addedge ( int u , int v ) {    E[cntE] = Edge ( v , H[u] ) ;    H[u] = cntE ++ ;}void dfs ( int u ) {    siz[u] = 1 ;    son[u] = 0 ;    for ( int i = H[u] ; ~i ; i = E[i].n ) {        int v = E[i].v ;        if ( v == pre[u] ) continue ;        pre[v] = u ;        dep[v] = dep[u] + 1 ;        dfs ( v ) ;        siz[u] += siz[v] ;        if ( siz[son[u]] < siz[v] ) son[u] = v ;    }    ou[u] = dfs_clock ;}void rebuild ( int u , int top_element ) {    in[u] = ++ dfs_clock ;    top[u] = top_element ;    pos[u] = ++ tree_idx ;    idx[tree_idx] = u ;    if ( son[u] ) rebuild ( son[u] , top_element ) ;    for ( int i = H[u] ; ~i ; i = E[i].n ) {        int v = E[i].v ;        if ( v != pre[u] && v != son[u] ) rebuild ( v , v ) ;    }    ou[u] = dfs_clock ;}void build ( int o , int l , int r ) {    tpos[o] = idx[l] ;    if ( l == r ) {        tree[l] = o ;        return ;    }    int m = mid ;    build ( lson ) ;    build ( rson ) ;}void update ( int x , int v ) {    int o = tree[x] ;    tpos[o] = v ? idx[x] : 0 ;    while ( o > 1 ) {        o >>= 1 ;        tpos[o] = max ( tpos[ls] , tpos[rs] ) ;    }}int query ( int L , int R , int o , int l , int r ) {    if ( L <= l && r <= R ) return tpos[o] ;    int m = mid ;    if ( R <= m ) return query ( L , R , lson ) ;    if ( m <  L ) return query ( L , R , rson ) ;    return max ( query ( L , R , lson ) , query ( L , R , rson ) ) ;}void Update1 ( int x , int y ) {    while ( top[x] != top[y] ) {        if ( dep[top[x]] < dep[top[y]] ) swap ( x , y ) ;        while ( 1 ) {            int t = query ( pos[top[x]] , pos[x] , root ) ;            if ( !t ) break ;            c[t] = 0 ;            update ( pos[t] , 0 ) ;            ans += val[t] ;            //printf ( "%d\n" , t ) ;        }        x = pre[top[x]] ;    }    if ( dep[x] > dep[y] ) swap ( x , y ) ;    while ( 1 ) {        int t = query ( pos[x] , pos[y] , root ) ;        if ( !t ) break ;        c[t] = 0 ;        update ( pos[t] , 0 ) ;        ans += val[t] ;    }}void Update2 ( int L , int R ) {    while ( 1 ) {        int t = query ( L , R , root ) ;        if ( !t ) break ;        c[t] = 0 ;        update ( pos[t] , 0 ) ;        ans += val[t] ;    }}void solve () {    int u , v , op ;    init () ;    scanf ( "%d" , &n ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        scanf ( "%d" , &val[i] ) ;        c[i] = 1 ;    }    for ( int i = 1 ; i < n ; ++ i ) {        scanf ( "%d%d" , &u , &v ) ;        addedge ( u , v ) ;        addedge ( v , u ) ;    }    dfs ( 1 ) ;    rebuild ( 1 , 1 ) ;    build ( root ) ;    scanf ( "%d" , &q ) ;    for ( int i = 0 ; i < q ; ++ i ) {        scanf ( "%d" , &op ) ;        if ( op == 1 ) {            scanf ( "%d%d" , &u , &v ) ;            Update1 ( u , v ) ;        } else if ( op == 2 ) {            scanf ( "%d" , &u ) ;            if ( !c[u] ) {                c[u] = 1 ;                update ( pos[u] , 1 ) ;                ans -= val[u] ;            }        } else {            scanf ( "%d" , &u ) ;            Update2 ( in[u] , ou[u] ) ;        }        printf ( "%d\n" , ans ) ;    }}int main () {    int T ;    scanf ( "%d" , &T ) ;    for ( int i = 1 ; i <= T ; ++ i ) {        solve () ;    }    return 0 ;}