题意：n个城市(n <= 10000)， 有m条边（m <= 40000），每一个城市有一个维护费用Cost（i），除此之外，每条边的维修费用为去掉该边后不能通信的城市对数与边权的积。这个费用要加到这条边的两端城市的某一个，问你全部城市的最大费用的最小值。、

思路：首先边的费用能够通过Tarjan求桥之后求得(利用桥的性质)，然后就是二分答案了！对于每一个点，假设有个儿子不能维护。那么不可行，否则。试着让儿子去维护边权，假设不可行，仅仅能让父亲承担。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 10000+10;
const int maxm = 40000+10;
LL cost[maxn];
int head[maxn];
int n,m;
int dfn[maxn],lown[maxn];
bool CanE[maxn];
int cnt[maxm];
LL ans;
int nume,dfs_clock;
bool vis[maxn];
bool isBridge[maxm];
vector<int> sta;
struct edge{
    int u,v,w,nxt;
}e[maxm];

void Tarjan(int u,int fa) {
    lown[u] = dfn[u] = dfs_clock++;
    sta.push_back(u);
    for(int i = head[u]; i ; i = e[i].nxt) {
        int v = e[i].v;
        if(v==fa) continue;
        if(!dfn[v]) {
            Tarjan(v,u);
            lown[u] = min(lown[u],lown[v]);
            if(lown[v] > dfn[u]) {
                isBridge[i] = isBridge[i^1] = true;
                cnt[i] = cnt[i^1] = dfs_clock-dfn[v];
            }
        }else {
            lown[u] = min(lown[u],dfn[v]);
        }
    }
}
bool dfs(int u,LL s,LL x) {
    vis[u] = true;
    LL ts = cost[u];
    int sz = sta.size();
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].v, w = e[i].w;
        if(isBridge[i] && !vis[v]) {
            LL tmp = (LL)w*cnt[i]*(sz-cnt[i]);
            if(!dfs(v,tmp,x)) return false;
            if(!CanE[v]) ts += tmp;
        }
    }
    if(ts > x) return false;
    if(ts+s <= x)  CanE[u] = true;
    return true;
}
bool can(LL x) {
    int len = sta.size();
    memset(CanE,false,sizeof CanE);
    for(int i = 0; i < len; i++) vis[sta[i]] = false;
    for(int i = 0; i < len; i++) {
        int t = sta[i];
        if(vis[t]) continue;
        if(!dfs(t,0,x)) return false;
    }
    return true;
}
void init() {
    nume = 1;
    ans = 0;
    dfs_clock = 1;
    memset(head,0,sizeof head);
    memset(isBridge,false,sizeof isBridge);
    memset(vis,false,sizeof vis);
    memset(dfn,0,sizeof dfn);
    memset(lown,0,sizeof lown);
    memset(CanE,false,sizeof CanE);
    sta.clear();
}
void addedge(int u,int v,int w) {
    e[++nume].nxt = head[u];
    e[nume].u = u;
    e[nume].v = v;
    e[nume].w = w;
    head[u] = nume;
}
void solve() {
    for(int i = 1; i <= n; i++) {
        if(!dfn[i]) {
            sta.clear();
            Tarjan(i,-1);
            LL L = ans,R = 1e10;
            while(L <= R) {
                LL mid = (L+R)/2;
                if(can(mid)) {
                    R = mid-1;
                }else{
                    L = mid+1;
                }
            }
            ans = max(ans,L);
        }
    }
    printf("%lld\n",ans);
}
int main(){

    int ncase,T=1;
    cin >> ncase;
    while(ncase--) {
        init();
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++) {
            scanf("%lld",&cost[i]);
            ans = max(ans,cost[i]);
        }
        for(int i = 0; i < m; i++) {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,c);
            addedge(b,a,c);
        }
        printf("Case %d: ",T++);
        solve();
    }
    return 0;
}