HDU - 6763 Total Eclipse

传送门：HDU - 6763

思路分析

从小到大复杂度太高了，考虑从大到小考虑，每遇到一个点，就加上这个点的贡献，然后遍历他的邻边，如果存在之前已经访问过的点，并且不在同一个集合，那么减去这个点的贡献

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 1e6+10;

int f[N],vis[N];

struct node{
    int id,w;
}a[N];

vector<int>e[N];

int find(int x){
    return x==f[x]?x:f[x]=find(f[x]);
}


int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    read(t);
    while(t--){
        int n,m;
        read(n,m);
        for(int i=1;i<=n;i++) {
            f[i]=i;
            vis[i]=0;
            e[i].clear();
            read(a[i].w);
            a[i].id=i;
        }
        while(m--){
            int u,v;
            read(u,v);
            e[u].pb(v);
            e[v].pb(u);
        }
        sort(a+1,a+1+n,[](node x,node y){
            return x.w>y.w;
        });
        ll ans=0;
        for(int i=1;i<=n;i++){
            int u=a[i].id;
            ans+=a[i].w;
            vis[u]=1;
            for(auto v:e[u]){
                if(vis[v]){
                    int fv=find(v);
                    if(u!=fv){
                        f[fv]=u;
                        ans-=a[i].w;
                    }
                }
            }
        }
        printf("%lld\n",ans);
    }




    return 0;
}