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 = 501000;

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=1;i<=k;i++){scanf("%d%d",&p1[i].from,&p1[i].to);}
    for(int i=1;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=0;i<g[now].size();i++){
    int to;int z = g[now][i];
    if(z > 0){
        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+1,now);
    }
}

void work(){
    for(int i=1;i<=n;i++) pre[i] = i;
    for(int i=1;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=1;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] = 1;
    }
    }
    dfs(1,1,0);
    for(int i=1;i<=n;i++) minn[i] = 1000000006,pre[i]=i;
    for(int i=1;i<=m;i++){
    if(arr[i] == 0){
        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!=0)pre[u] = z; minn[fd[u]] = p2[i].w; u = z;
        }else{
            int z = found(fa[v]);
            if(z!=0)pre[v] = z; minn[fd[v]] = p2[i].w; v = z;
        }
        }
    }
    }
    long long ans = 0;
    for(int i=1;i<=k;i++){
    if(minn[i] == 1000000006) {puts("-1");return;}
    else ans += minn[i];
    }
    printf("%I64d",ans);
}

int main(){
    read();
    work();
    return 0;
}