杭电多校第九场-1005-Resistance

既然开始水博客了,就多水点吧……

你有一颗仙人掌(每两个环至多有一个交点)形状的电路,一条边的电阻是1。你需要求出所有点对之间的电阻的和,对1e9+7取模。

第一眼看到“仙人掌”,只能想到某些仙人掌数据结构。再仔细一看,才发现挺妙的……

OCR实在是太强大了,我直接复制了出题人的题解,反正就是这么做的。注意dfs树的性质,所有不是树边的边都是返祖边,可以写起来很简单。据说fft可能tle,但是我还是一遍过了,可能是kactl的模版实在太快了吧,牛逼。。

出题人的题解如下

考虑两点之间的电阻,首先你把路径拿出来,可以看成若干条树边和环串联起来,也就是相加。所以考虑也就是对于每个树边和环分别考虑对答案的贡献。
每个树边的贡献为通过树边的点对,也就是两边大小相乘。对于环边,假设两个点下面的子树大小分别 为 S_{i}S_{j}, 环长为l,距离为 j-i, 那么等效电阻为 j-il-j+i 并联,也就是 (l-j+i)(j-i) / l 共有 S_{i} S_{j} 对,可以将式子展开之后使用前缀和之类的方法计算,使用FFT可能会TLE。。

代码如下

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

template <typename A, typename B>string to_string(pair<A, B> p);template <typename A, typename B, typename C>string to_string(tuple<A, B, C> p);template <typename A, typename B, typename C, typename D>string to_string(tuple<A, B, C, D> p);string to_string(const string& s) {  return '"' + s + '"';}string to_string(const char* s) {  return to_string((string) s);}string to_string(bool b) {  return (b ? "true" : "false");}string to_string(vector<bool> v) {  bool first = true;  string res = "{";  for (int i = 0; i < static_cast<int>(v.size()); i++) {    if (!first) {      res += ", ";    }    first = false;    res += to_string(v[i]);  }  res += "}";  return res;}template <size_t N>string to_string(bitset<N> v) {  string res = "";  for (size_t i = 0; i < N; i++) {    res += static_cast<char>('0' + v[i]);  }  return res;}template <typename A>string to_string(A v) {  bool first = true;  string res = "{";  for (const auto &x : v) {    if (!first) {      res += ", ";    }    first = false;    res += to_string(x);  }  res += "}";  return res;}template <typename A, typename B>string to_string(pair<A, B> p) {  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";}template <typename A, typename B, typename C>string to_string(tuple<A, B, C> p) {  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";}template <typename A, typename B, typename C, typename D>string to_string(tuple<A, B, C, D> p) {  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";}void debug_out() { cerr << endl; }template <typename Head, typename... Tail>void debug_out(Head H, Tail... T) {  cerr << " " << to_string(H);  debug_out(T...);}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define set0(x) memset(x,0,sizeof(x))
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<typename T> void read(T &x){
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
    read(first);
    read(args...);
}

const long double M_PIl = acosl(-1);
typedef complex<double> C;
typedef vector<double> vd;
typedef vector<int> vi;
void fft(vector<C>& a) {
    int n = sz(a), L = 31 - __builtin_clz(n);
    static vector<complex<long double>> R(2, 1);
    static vector<C> rt(2, 1);  // (^ 10% faster if double)
    for (static int k = 2; k < n; k *= 2) {
        R.resize(n); rt.resize(n);
        auto x = polar(1.0L, M_PIl / k); // M_PI, lower-case L
        rep(i,k,2*k) rt[i] = R[i] = i&1 ? R[i/2] * x : R[i/2];
    }
    vi rev(n);
    rep(i,0,n) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
    rep(i,0,n) if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int k = 1; k < n; k *= 2)
        for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
            // C z = rt[j+k] * a[i+j+k]; // (25% faster if hand-rolled)  /// include-line
            auto x = (double *)&rt[j+k], y = (double *)&a[i+j+k];        /// exclude-line
            C z(x[0]*y[0] - x[1]*y[1], x[0]*y[1] + x[1]*y[0]);           /// exclude-line
            a[i + j + k] = a[i + j] - z;
            a[i + j] += z;
        }
}
vd conv(const vd& a, const vd& b) {
    if (a.empty() || b.empty()) return {};
    vd res(sz(a) + sz(b) - 1);
    int L = 32 - __builtin_clz(sz(res)), n = 1 << L;
    vector<C> in(n), out(n);
    copy(all(a), begin(in));
    rep(i,0,sz(b)) in[i].imag(b[i]);
    fft(in);
    trav(x, in) x *= x;
    rep(i,0,n) out[i] = in[-i & (n - 1)] - conj(in[i]);
    fft(out);
    rep(i,0,sz(res)) res[i] = imag(out[i]) / (4 * n);
    return res;
}

const int mod = 1000000007;
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline int sub(int x,int y){return x-y<0?x-y+mod:x-y;}
inline int sq(int x){return 1ll*x*x%mod;}
int mpow(int a,int b){return b == 0 ? 1 : ( b&1 ? mul(a,sq(mpow(a,b/2))) : sq(mpow(a,b/2)));}

int T,n,m,u,v,vis[200020],depth[200020],sz[200020],fa[200020],ok[200020];
vector<int> G[200020];

int ans = 0;

void dfs(int num){
    debug(num);
    vis[num] = 1;
    sz[num] = 1;
    for(auto ct:G[num]){
        if(!vis[ct]){
            depth[ct] = depth[num]+1;
            fa[ct] = num;
            dfs(ct);
            sz[num]+=sz[ct];
        }else{
            if(depth[ct] == depth[num]-1 || depth[ct] > depth[num]+1)continue;
            debug(num,ct);
            int p = num;
            while(p!=ct){
                debug(p);
                ok[p] = 0;
                p = fa[p];
            }
        }
    }
}

void dfs2(int num){
    debug(num);
    if(ok[num] && num!=1){
        ans = add(ans,mul(sz[num],n-sz[num]));
    }
    for(auto ct:G[num]){
        if(depth[ct] == depth[num]+1)dfs2(ct);
        else{
            if(depth[ct] == depth[num]-1 || depth[ct] > depth[num]+1)continue;
            debug(num,ct,depth[num],depth[ct]);
            int circsize = depth[num]-depth[ct]+1;
            vector<int> val;
            val.PB(sz[num]);
            int v1 = fa[num],v2 = num;
            while(v2!=ct){
                val.PB(sz[v1]-sz[v2]);
                v2 = v1;
                v1 = fa[v1];
            }
            val[val.size()-1]+=n-sz[ct];
            vd a,b;
            for(auto ct:val)a.PB(ct);
            b = a;
            reverse(all(b));
            a = conv(a,b);
            int cv = 0;
            for(int i=0;i<a.size();i++){
                ll c = ((ll)(a[i]+0.5))%mod;
                int d = abs(i-(circsize-1));
                cv = add(cv,mul(c,mul(d,circsize-d)));
            }
            cv = mul(cv,(mod+1)/2);
            cv = mul(cv,mpow(circsize,mod-2));
            ans = add(ans,cv);
        }
    }
}

int main() {
    read(T);
    while(T--){
        ans = 0;
        read(n,m);
        fa[1] = 1;
        for(int i=1;i<=n;i++){
            vis[i] = 0;
            depth[i] = 0;
            ok[i] = 1;
            G[i].clear();
        }
        for(int i=1;i<=m;i++){
            read(u,v);
            G[u].PB(v);
            G[v].PB(u);
        }
        debug(1);
        dfs(1);
        debug(1);
        dfs2(1);
        cout<<ans<<endl;
    }
    return 0;
}

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