HDU - 6832 A Very Easy Graph Problem

传送门：HDU - 6832

思路分析

因为边是指数型增长的，每加一条边，比之前所有边加起来都多，所以如果两个点之前已经连通了，这条边就没贡献了，用并查集维护一下最小生成树，然后就是简单的树DP了

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 = 1e5+10;
const int mod = 1e9+7;

#define int long long

int n,m,f[N];
vector<pair<int,int>>e[N];
ll dis[N][2],num[N][2],ans[N],a[N];

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


void dfs(int u,int fa){
    num[u][a[u]]++;
    for(auto x:e[u]){
        int v=x.fi;
        int w=x.se;
        if(v==fa) continue;
        dfs(v,u);
        num[u][0]+=num[v][0];
        num[u][1]+=num[v][1];
        dis[u][0]=(dis[u][0] + dis[v][0] + num[v][0]*w%mod)%mod;
        dis[u][1]=(dis[u][1] + dis[v][1] + num[v][1]*w%mod)%mod;
        ans[u] = (ans[u]+ans[v]) %mod;
    }
    for(auto x:e[u]){
        int v=x.fi;
        int w=x.se;
        if(v==fa) continue;
        ans[u] = (ans[u] + (num[u][0] - num[v][0]) * (dis[v][1] + num[v][1]*w)%mod ) %mod;
        ans[u] = (ans[u] + (num[u][1] - num[v][1]) * (dis[v][0] + num[v][0]*w)%mod ) %mod;
    }
}

signed main() {

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

    int T;
    read(T);
    while(T--) {
        read(n,m);
        for(int i=1; i<=n; i++) {
            read(a[i]);
            ans[i]=0;
            e[i].clear();
            num[i][0]=num[i][1]=0;
            dis[i][0]=dis[i][1]=0;
            f[i]=i;
        }
        for(int i=1,w=2; i<=m; i++) {
            int u,v;
            read(u,v);
            int fu=find(u);
            int fv=find(v);
            if(fu!=fv) {
                e[u].pb(mp(v,w));
                e[v].pb(mp(u,w));
                f[fu]=fv;
            }
            w=w*2%mod;
        }
        dfs(1,1);
        
        printf("%lld\n",ans[1]);
    }




    return 0;
}