【HDU】5016 Mart Master II 点分治

传送门：【HDU】5016 Mart Master II

题目分析：终于写出来了。。。1A。。

首先我们可以预处理出每个不是mart的点x所属与的mart的编号idx[x]以及到该mart的距离d[x]，用一个结构体表示。

该部分最短路算法即可。

我们定义结构体变量x小于y即d[x]<d[y] ||d[x] == d[y]&&idx[x]<idx[y]。

然后便是树分治的舞台了~

首先找到树的重心，然后以该重心dfs一次得到所有点x到重心的距离dis[x]以及这棵树上的节点总数cnt，将d[x]-dis[x]以及idx[x]组成结构体丢进数组S，dfs以后对数组S排序，然后二分小于等于dis[x]的数目tmp，则该子树上能被x占领的数量为cnt-tmp。

为什么是cnt-tmp呢？因为如果x要占领y，则dis[x]+dis[y]<d[y]才行，所以有dis[x]<d[y]-dis[y]，因为dis[x]>=d[y]-dis[y]的数目即tmp，节点总数即cnt，所以子树内的贡献为cnt-tmp。

并且，由于可能有的贡献是来自同一棵子树的，我们需要将其去重。

接下来就是递归求解了。

具体部分可以见代码，代码还是比较清晰的。

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 Log( i , a , b ) for ( int i = a ; ( 1 << i ) <= b ; ++ i )#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 )const int MAXN = 100005 ;const int MAXE = 200005 ;const int INF = 0x3f3f3f3f ;struct Edge {int v , c ;Edge* next ;} E[MAXE] , *H[MAXN] , *edge ;struct Node {int dis ;int idx ;Node () {}Node ( int dis , int idx ) : dis ( dis ) , idx ( idx ) {}bool operator < ( const Node& a ) const {if ( dis != a.dis ) return dis < a.dis ;return idx < a.idx ;}bool operator > ( const Node& a ) const {return a < *this ;}bool operator <= ( const Node& a ) const {return !( a < *this ) ;}bool operator >= ( const Node& a ) const {return !( *this < a ) ;}Node operator + ( const int& val ) const {return Node ( dis + val , idx ) ;}Node operator - ( const int& val ) const {return Node ( dis - val , idx ) ;}} node[MAXN] , S[MAXN] ;int Q[MAXN] , head , tail ;int vis[MAXN] ;int Time ;int siz[MAXN] ;int num[MAXN] ;int mart[MAXN] ;int ans[MAXN] ;int dis[MAXN] ;int size ;int root ;int cnt ;int n ;void clear () {edge = E ;num[0] = n ;clr ( H , 0 ) ;head = tail = 0 ;clr ( ans , 0 ) ;}void addedge ( int u , int v , int c ) {edge->v = v ;edge->c = c ;edge->next = H[u] ;H[u] = edge ++ ;}void spfa () {while ( head != tail ) {int u = Q[head ++] ;if ( head == MAXN ) head = 0 ;vis[u] = Time - 1 ;travel ( e , H , u ) {int v = e->v ;Node tmp = node[u] + e->c ;if ( node[v] > tmp ) {node[v] = tmp ;if ( vis[v] != Time ) {vis[v] = Time ;Q[tail ++] = v ;if ( tail == MAXN ) tail = 0 ;}}}}}void get_siz ( int u , int fa = 0 ) {siz[u] = 1 ;travel ( e , H , u ) {int v = e->v ;if ( v != fa && vis[v] != Time ) {get_siz ( v , u ) ;siz[u] += siz[v] ;}}}void get_root ( int u , int fa = 0 ) {num[u] = 0 ;travel ( e , H , u ) {int v = e->v ;if ( v != fa && vis[v] != Time ) {get_root ( v , u ) ;num[u] = max ( num[u] , siz[v] ) ;}}num[u] = max ( num[u] , size - siz[u] ) ;if ( num[u] < num[root] ) root = u ;}void get_dis ( int u , int val , int fa = 0 ) {dis[u] = val ;S[++ cnt] = node[u] - val ;//丢进S数组travel ( e , H , u ) {int v = e->v ;if ( v != fa && vis[v] != Time ) {get_dis ( v , dis[u] + e->c , u ) ;}}}int search ( const Node& x ) {//二分查找，如果没有找到目标，则返回小于x的最大的数的下标int l = 0 , r = cnt ;while ( l < r ) {int m = ( l + r + 1 ) >> 1 ;if ( S[m] <= x ) l = m ;else r = m - 1 ;}return r ;}void get_ans ( int u , int sign , int fa = 0 ) {if ( !mart[u] ) {//为不是mart的点计算能被其占领的点的个数int tmp = search ( Node ( dis[u] , u ) ) ;ans[u] += ( cnt - tmp ) * sign ;}travel ( e , H , u ) {int v = e->v ;if ( v != fa && vis[v] != Time ) {get_ans ( v , sign , u ) ;}}}void deal ( int u , int val , int sign ) {cnt = 0 ;get_dis ( u , val ) ;//得到distsort ( S + 1 , S + cnt + 1 ) ;get_ans ( u , sign ) ;//计算贡献}void divide ( int u ) {//分治get_siz ( u ) ;//得到子树规模size = siz[u] ;root = 0 ;get_root ( u ) ;//寻找u所在的子树的重心vis[root] = Time ;deal ( root , 0 , 1 ) ;//得到该子树内所有的贡献，不管是不是同一棵子树的travel ( e , H , root ) if ( vis[e->v] != Time ) deal ( e->v , e->c , -1 ) ;//去重，排除同一棵子树的贡献travel ( e , H , root ) if ( vis[e->v] != Time ) divide ( e->v ) ;//递归处理}void solve () {int x , y , c ;clear () ;rep ( i , 1 , n ) {scanf ( "%d%d%d" , &x , &y , &c ) ;addedge ( x , y , c ) ;addedge ( y , x , c ) ;}For ( i , 1 , n ) {scanf ( "%d" , &mart[i] ) ;if ( mart[i] ) {node[i] = Node ( 0 , i ) ;Q[tail ++] = i ;} else node[i] = Node ( INF , 0 ) ;}++ Time ;spfa () ;++ Time ;divide ( 1 ) ;int res = 0 ;For ( i , 1 , n ) if ( ans[i] > res ) res = ans[i] ;printf ( "%d\n" , res ) ;}int main () {Time = 0 ;clr ( vis , 0 ) ;//优化while ( ~scanf ( "%d" , &n ) ) solve () ;return 0 ;}