Codeforces1023F Mobile Phone Network 【并查集】【最小生成树】

题目大意:

给一些没安排权值的边和安排了权值的边，没被安排的边全要被选入最小生成树，问你最大能把它们的权值和安排成多少。
题目分析:
假设建好了树，那么树边与剩下的每一条边都能构成一个环，并且非树边的权值是环中最大的，所以钦定边权不大于非树边即可。用并查集维护一下。

代码:

 #include<bits/stdc++.h>
 using namespace std;

 const int maxn = ;

 int n,k,m;
 struct edge{int from,to,w;}p1[maxn],p2[maxn];
 int pre[maxn],arr[maxn],minn[maxn],dep[maxn],fa[maxn],fd[maxn];
 vector<int> g[maxn];

 int found(int x){
     int rx = x; while(pre[rx] != rx) rx = pre[rx];
     while(pre[x] != rx){int tmp = pre[x]; pre[x] = rx; x = tmp;}
     return rx;
 }

 void read(){
     scanf("%d%d%d",&n,&k,&m);
     for(int i=;i<=k;i++){scanf("%d%d",&p1[i].from,&p1[i].to);}
     for(int i=;i<=m;i++){scanf("%d%d%d",&p2[i].from,&p2[i].to,&p2[i].w);}
 }

 void dfs(int now,int dp,int f){
     dep[now] = dp; fa[now] = f;
     for(int i=;i<g[now].size();i++){
     int to;int z = g[now][i];
     if(z > ){
         if(p1[z].to == now) to = p1[z].from;
         else to = p1[z].to;
     }else{
         if(p2[-z].to == now) to = p2[-z].from;
         else to = p2[-z].to;
     }
     if(to == f) continue;
     if(z) fd[to] = z;
     dfs(to,dp+,now);
     }
 }

 void work(){
     for(int i=;i<=n;i++) pre[i] = i;
     for(int i=;i<=k;i++){
     if(found(p1[i].from) != found(p1[i].to)){
         pre[found(p1[i].from)] = found(p1[i].to);
         g[p1[i].from].push_back(i); g[p1[i].to].push_back(i);
     }
     }
     for(int i=;i<=m;i++){
     if(found(p2[i].from) != found(p2[i].to)){
         pre[found(p2[i].from)] = found(p2[i].to);
         g[p2[i].from].push_back(-i); g[p2[i].to].push_back(-i);
         arr[i] = ;
     }
     }
     dfs(,,);
     for(int i=;i<=n;i++) minn[i] = ,pre[i]=i;
     for(int i=;i<=m;i++){
     if(arr[i] == ){
         int u = found(p2[i].from),v = found(p2[i].to);
         while(u != v){
         if(dep[u] >= dep[v]){
             int z = found(fa[u]);
             if(z!=)pre[u] = z; minn[fd[u]] = p2[i].w; u = z;
         }else{
             int z = found(fa[v]);
             if(z!=)pre[v] = z; minn[fd[v]] = p2[i].w; v = z;
         }
         }
     }
     }
     long long ans = ;
     for(int i=;i<=k;i++){
     if(minn[i] == ) {puts("-1");return;}
     else ans += minn[i];
     }
     printf("%I64d",ans);
 }

 int main(){
     read();
     work();
     return ;
 }