1011 Highway 树的直径 树的最大生成树
链接：https://ac.nowcoder.com/acm/contest/26077/1011
来源：牛客网
题目描述
In ICPCCamp there were n towns conveniently numbered with 1,2,…,n1, 2, \dots, n1,2,…,n
connected with (n - 1) roads.
The i-th road connecting towns aia_iai and bib_ibi has length cic_ici.
It is guaranteed that any two cities reach each other using only roads.
Bobo would like to build (n - 1) highways so that any two towns reach each using *only highways*.
Building a highway between towns x and y costs him δ(x,y)\delta(x, y)δ(x,y) cents,
where δ(x,y)\delta(x, y)δ(x,y) is the length of the shortest path between towns x and y using roads.
As Bobo is rich, he would like to find the most expensive way to build the (n - 1) highways.
connected with (n - 1) roads.
The i-th road connecting towns aia_iai and bib_ibi has length cic_ici.
It is guaranteed that any two cities reach each other using only roads.
Bobo would like to build (n - 1) highways so that any two towns reach each using *only highways*.
Building a highway between towns x and y costs him δ(x,y)\delta(x, y)δ(x,y) cents,
where δ(x,y)\delta(x, y)δ(x,y) is the length of the shortest path between towns x and y using roads.
As Bobo is rich, he would like to find the most expensive way to build the (n - 1) highways.
输入描述:
The input contains zero or more test cases and is terminated by end-of-file. For each test case:
The first line contains an integer n.
The i-th of the following (n - 1) lines contains three integers aia_iai, bib_ibi and cic_ici.
* 1≤n≤1051 \leq n \leq 10^51≤n≤105
* 1≤ai,bi≤n1 \leq a_i, b_i \leq n1≤ai,bi≤n
* 1≤ci≤1081 \leq c_i \leq 10^81≤ci≤108
* The number of test cases does not exceed 10.
输出描述:
For each test case, output an integer which denotes the result.
分析
要求的是根据所给的n - 1条边求最大生成树。
首先要知道，任何一个点，在这颗树上，只有连向这颗树的直径上的两个点u1，u2的路径才是最长的。
不妨假设树的直径只有两个点u1，u2。
直径之间先连接一条边，再让剩下n - 2个点按照向哪个连接路径长度长就连哪个的规律去连接。就可以生成一个n - 1条边的最大生成树了。
//-------------------------代码---------------------------- #define int ll const int N = 1e5+10; int n,m; int dist1[N],dist2[N],c; V<pii> e[N]; void dfs(int u,int fa,int *dist) { for(auto it:e[u]) { if(it.x == fa) continue; dist[it.x] = dist[u] + it.y; if(dist[it.x] > dist[c]) c = it.x; dfs(it.x,u,dist); } } void solve() { // cin>>n>>m; ms(dist1,0);ms(dist2,0); fo(i,0,N-1) e[i].clear(); fo(i,1,n-1) { int a,b,c;cin>>a>>b>>c; e[a].pb({b,c}); e[b].pb({a,c}); } dfs(1,0,dist1); int u1 = c; ms(dist1,0); dfs(c,0,dist1); int u2 = c; dfs(c,0,dist2); int d = 0; fo(i,1,n) { d += max(dist1[i],dist2[i]); } d -= dist2[c]; cout<<d<<endl; } void main_init() {} signed main(){ AC();clapping();TLE; cout<<fixed<<setprecision(12); main_init(); while(cin>>n) // while(cin>>n>>m,n,m) // int t;cin>>t;while(t -- ) solve(); // {solve(); } return 0; } /*样例区 */ //------------------------------------------------------------
