The wheel of the history rolling forward, our king conquered a new region in a distant continent.

There are N towns (numbered from 1 to N) in this region connected by several roads. It's confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route. 

Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.

题意：有若干个城市，城市之间有边，边有容量，现在要求一个城市，使它到所有城市的总运量最大，到某一城市的运量等于相互路径上的最小容量。

由于运量是由最小容量限制的，所以我们可以先构建一颗最大生成树，边构建树边计算权值。将边从大到小排序之后，每次将最大权的边进行合并，合并的两个块内部的边都是在之前合并的，因此边权一定比当前合并的边权大，所以这两块之间运输的相互运量均等于当前边的边权，这样就可以边合并边计算合并后的最大运量，即等于某一边的最大运量加上边权乘另一边的点数个数，两边取最大值即为合并后的最大权值。

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 using namespace std; 6 typedef long long ll; 7 const int INF=0x3f3f3f3f; 8 const int mod=1e9+7; 9 const int maxn=2e6+5;10 11 struct seg{12     int a,b;13     ll v;14     bool operator <(const seg x)const{15         return v>x.v;16     }17 }s[maxn];18 19 int fa[maxn];20 ll sz[maxn];21 ll ma[maxn],ans;22 int n;23 24 int find(int x){
     return fa[x]==x?x:fa[x]=find(fa[x]);}25 26 void unio(int i){27     int x=s[i].a,y=s[i].b;28     ll v=s[i].v;29     int X=find(x),Y=find(y);30     ll nx=ma[X]+v*sz[Y];31     ll ny=ma[Y]+v*sz[X];32     if(nx>=ny){33         fa[Y]=X;34         sz[X]+=sz[Y];35         ma[X]=nx;36         if(nx>ans)ans=nx;37     }38     else{39         fa[X]=Y;40         sz[Y]+=sz[X];41         ma[Y]=ny;42         if(ny>ans)ans=ny;43     }44 }45 46 void init(){47     ans=0;48     for(int i=0;i<=n+5;++i)fa[i]=i;49     for(int i=0;i<=n+5;++i)sz[i]=1;50     for(int i=0;i<=n+5;++i)ma[i]=0;51 }52 53 int main(){54     while(scanf("%d",&n)!=EOF){55         if(n==1){printf("0\n");continue;}56         init();57         for(int i=1;i