本文主要是介绍【HDU】4871 Shortest-path tree 最短路+点分治，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
传送门：【HDU】4871 Shortest-path tree
题目分析：
学了点分治后来看这道题，简直就是水题。。。
但是我竟然花了将近一个晚上才写出来。。。就因为一个地方写漏了T U T。。
首先根据题意求一棵树，最短路一下，然后最小字典序就按照编号顺序遍历邻接表给节点标记。
剩下的就是树分治的事了。
在以重心X为根的子树中，按照X的子节点v的子树中最长路径包含节点数升序遍历X的子节点v，每次对v做dfs求得v上每个结点数目的最长长度以及数量，用dp[k],cnt[k]表示（dp[k]为结点数为k的路径的最长长度，cnt[k]为对应的路径数）。然后用F[k]表示v之前的子树中结点数为k的最长长度，G[k]为对应的路径数。则遍历所有子树v可能的结点数，更新结点数为K的最长长度以及数量，之后再更新F[k]以及G[k]。
最后输出答案即可。
代码写的十分丑陋，勉强看看好了。。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std ;typedef long long LL ;#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 = 30005 ;
const int MAXE = 200005 ;
const int INF = 0x3f3f3f3f ;struct Edge {int v , c ;Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;struct Node {int num , v , c ;Node () {}Node ( int num , int v , int c ) : num ( num ) , v ( v ) , c ( c ) {}bool operator < ( const Node& a ) const {return num < a.num ;}
} T[MAXN] ;struct Shortest_Path {Edge E[MAXE] , *H[MAXN] , *edge ;int Q[MAXN] , head , tail ;int d[MAXN] ;bool vis[MAXN] ;int n ;void clear () {edge = E ;clr ( H , 0 ) ;}void Addedge ( int u , int v , int c ) {edge -> v = v ;edge -> c = c ;edge -> next = H[u] ;H[u] = edge ++ ;}void spfa ( int s ) {clr ( d , INF ) ;clr ( vis , 0 ) ;head = tail = 0 ;d[s] = 0 ;Q[tail ++] = s ;while ( head != tail ) {int u = Q[head ++] ;if ( head == MAXN ) head = 0 ;vis[u] = 0 ;travel ( e , H , u ) {int v = e -> v , c = e -> 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 ;}}}}}}void build_tree () ;
} S ;bool vis[MAXN] ;
int siz[MAXN] ;
int num[MAXN] ;
int cnt[MAXN] ;
int dp[MAXN] ;
int G[MAXN] ;
int F[MAXN] ;int ans1 , ans2 ;
int node_num ;
int n , K ;
int root ;void clear () {S.n = n ;edge = E ;num[0] = n ;S.clear () ;clr ( H , 0 ) ;clr ( vis , 0 ) ;ans1 = ans2 = 0 ;
}void addedge ( int u , int v , int c ) {edge -> v = v ;edge -> c = c ;edge -> next = H[u] ;H[u] = edge ++ ;
}void get_siz ( int u , int fa = 0 ) {siz[u] = 1 ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] && v != fa ) {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 ( !vis[v] && v != fa ) {get_root ( v , u ) ;num[u] = max ( num[u] , siz[v] ) ;}}num[u] = max ( num[u] , node_num - siz[u] ) ;if ( num[u] < num[root] ) root = u ;
}int get_num ( int u , int fa = 0 , int d = 0 ) {int res = 0 ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] && v != fa ) res = max ( res , get_num ( v , u , d + 1 ) ) ;}return res + 1 ;
}void get_cnt ( int u , int fa , int k , int val ) {if ( k > K ) return ;if ( val > dp[k] ) {dp[k] = val ;cnt[k] = 1 ;} else if ( val == dp[k] ) cnt[k] ++ ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] && v != fa ) get_cnt ( v , u , k + 1 , val + e -> c ) ;}
}void tree_dfs ( int u ) {get_siz ( u ) ;root = 0 ;node_num = siz[u] ;get_root ( u ) ;if ( num[root] * 2 + 1 < K ) return ;//优化int rt = root , m = 0 ;vis[rt] = 1 ;travel ( e , H , rt ) {int v = e -> v ;if ( !vis[v] ) tree_dfs ( v ) ;}travel ( e , H , rt ) {int v = e -> v , c = e -> c ;if ( !vis[v] ) T[m ++] = Node ( get_num ( v ) , v , c ) ;}sort ( T , T + m ) ;FOR ( i , 0 , K ) F[i] = -INF , G[i] = 0 ;//rep ( i , 0 , m ) {FOR ( j , 1 , T[i].num ) dp[j] = -INF , cnt[j] = 0 ;get_cnt ( T[i].v , rt , 1 , T[i].c ) ;int N = min ( K , T[i].num ) ;FOR ( j , 1 , N ) {if ( j == K - 1 ) {if ( dp[j] > ans1 ) {ans1 = dp[j] ;ans2 = cnt[j] ;} else if ( dp[j] == ans1 ) ans2 += cnt[j] ;break ;}int tmp = K - j - 1 ;if ( dp[j] + F[tmp] > ans1 ) {ans1 = dp[j] + F[tmp] ;ans2 = cnt[j] * G[tmp] ;} else if ( dp[j] + F[tmp] == ans1 ) ans2 += cnt[j] * G[tmp] ;}FOR ( j , 1 , N ) {if ( dp[j] > F[j] ) {F[j] = dp[j] ;G[j] = cnt[j] ;} else if ( dp[j] == F[j] ) G[j] += cnt[j] ;}}vis[rt] = 0 ;
}void Shortest_Path :: build_tree () {clr ( vis , 0 ) ;vis[1] = 1 ;FOR ( u , 1 , n ) {travel ( e , H , u ) {int v = e -> v , c = e -> c ;if ( !vis[v] && d[v] == d[u] + c ) {addedge ( u , v , c ) ;addedge ( v , u , c ) ;vis[v] = 1 ;}}}
}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 , c , m ;scanf ( n ) ; scanf ( m ) ; scanf ( K ) ;clear () ;rep ( i , 0 , m ) {scanf ( u ) , scanf ( v ) , scanf ( c ) ;S.Addedge ( u , v , c ) ;S.Addedge ( v , u , c ) ;}S.spfa ( 1 ) ;S.build_tree () ;tree_dfs ( 1 ) ;printf ( "%d %d\n" , ans1 , ans2 ) ;
}int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;
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