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 <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 , nconst 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：
一个人的期望值越大，心理承受力就会越小，就越经受不住失败的打击，
