hdu 5886 Tower Defence 2016ACM/ICPC青岛赛区网络赛1009

Problem Description

There was a civil war between two factions in Skyrim, a province of the Empire on the continent of Tamriel. The Stormcloaks, led by Ulfric Stormcloak, are made up of Skyrim's native Nord race. Their goal is an independent Skyrim free from Imperial interference. The Imperial Legion, led by General Tullius, is the military of the Empire that opposes the Stormcloaks and seeks to reunite and pacify the province.

The current target of General Tullius is to defend Whiterun City. Near by this city there are N towers under the Empire's control. There are N−1 roads link these tower, so solders can move from any tower to another one through these roads.

In military affairs, tactical depth means the longest path between two towers of all. Larger the tactical depth is, more stable these towers are.

According to the message sent by spies, General Tullius believe that Stormcloaks is planning to attack one of these roads, and his towers would be divided into two parts. However, Tullius does not know which one, so he supposes the possibility that Stormcloaks attack these roads are the same. Now, General Tullius ask for your help, to calculate the expectation of tactical depth after this attack.

To avoid the issue of precision, you need to calculate expectationoftacticaldepth×(N−1).

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
For each test case, in the first line there is an integer N(N≤100000).
The i-th line of the next N−1 lines describes the i-th edge. Three integers u,v,w (0≤w≤1000) describe an edge between u and v of length w.

 

Output

For each test cases, output expectationoftacticaldepth×(N−1).

 

Sample Input

232 1 23 2 552 1 73 1 74 2 55 2 6

 

Sample Output

763

给你一棵树，枚举删除每一条边，求剩下的直径之和

首先如果删除的边不在直径上，那贡献就是直径

如果删除的在直径上。我们假设一条直径的起点是L终点是R

那么我们分别以L和R为根处理每个子树的内部最大直径

然后统计一下就可以了

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct line{int s,t;int x;int next;}a[200001];int head[100001];int edge;inline void add(int s,int t,int x){a[edge].next=head[s];head[s]=edge;a[edge].s=s;a[edge].t=t;a[edge].x=x;}int fa[100001];long long son1[100001],son2[100001];int sonx1[100001],sonx2[100001];int dep[100001][3];inline void dfs1(int d){son1[d]=0;sonx1[d]=0;son2[d]=0;sonx2[d]=0;int i;for(i=head[d];i!=0;i=a[i].next){int t=a[i].t;if(t!=fa[d]){fa[t]=d;dfs1(t);if(son1[t]+a[i].x>=son1[d]){son2[d]=son1[d];sonx2[d]=sonx1[d];son1[d]=son1[t]+a[i].x;sonx1[d]=t;}else if(son1[t]+a[i].x>=son2[d]){son2[d]=son1[t]+a[i].x;sonx2[d]=t;}}}}bool v[100001];bool vx[100001];inline int findline(int d,int x){vx[d]=true;if(sonx1[d]==0)return d;dep[sonx1[d]][x]=dep[d][x]+1;return findline(sonx1[d],x);}long long ans[3][100001];inline void trdp(int d,int x){son1[d]=0;sonx1[d]=0;son2[d]=0;sonx2[d]=0;v[d]=true;int i;long long maxx=0;for(i=head[d];i!=0;i=a[i].next){int t=a[i].t;if(!v[t]){trdp(t,x);maxx=max(maxx,ans[x][t]);if(son1[t]+a[i].x>=son1[d]){son2[d]=son1[d];sonx2[d]=sonx1[d];son1[d]=son1[t]+a[i].x;sonx1[d]=t;}else if(son1[t]+a[i].x>=son2[d]){son2[d]=son1[t]+a[i].x;sonx2[d]=t;}}}ans[x][d]=max(son1[d]+son2[d],maxx);}int main(){//    freopen("1009.in", "r", stdin);//    freopen("1009.out", "w", stdout);int T;scanf("%d",&T);while(T>0){T--;edge=0;memset(head,0,sizeof(head)); int n;scanf("%d",&n);int i;int s,t,x;for(i=1;i<=n-1;i++){scanf("%d%d%d",&s,&t,&x);edge++;add(s,t,x);edge++;add(t,s,x);}memset(fa,0,sizeof(fa));dfs1(1);long long maxx=0,maxi=0;for(i=1;i<=n;i++){if(son1[i]+son2[i]>maxx){maxx=son1[i]+son2[i];maxi=i;}}int l,r;memset(dep,0,sizeof(dep));memset(vx,false,sizeof(vx));dep[sonx1[maxi]][1]=1;dep[sonx2[maxi]][2]=1;vx[maxi]=true;l=findline(sonx1[maxi],1);r=findline(sonx2[maxi],2);if(r==0)r=maxi;memset(ans,0,sizeof(ans));memset(v,false,sizeof(v));trdp(l,1);memset(v,false,sizeof(v));trdp(r,2);long long ansx=0;for(i=1;i<=edge;i+=2){s=a[i].s;t=a[i].t;if(vx[s]&&vx[t]){if(dep[s][1]<dep[t][1])ansx+=max(ans[1][s],ans[2][t]);else if(dep[s][1]>dep[t][1])ansx+=max(ans[2][s],ans[1][t]);else if(dep[s][2]<dep[t][2])ansx+=max(ans[2][s],ans[1][t]);else if(dep[s][2]>dep[t][2])ansx+=max(ans[1][s],ans[2][t]);}elseansx+=maxx;}printf("%I64d\n",ansx);}return 0;}