Codeforces 840D Expected diameter of a tree 分块思想

我们先两次dfs计算出每个点能到达最远点的距离。

暴力计算两棵树x, y连边直径的期望很好求，我们假设SZ(x) < SZ(y)

我们枚举 x 的每个端点，二分找到分界点，复杂度为SZ(x) * log(SZ(y))

其实我们对于每次询问我们记忆化一下就可以啦。

这是因为对于SZ(x)小于 sqrt(n)的询问，我们直接暴力求就好啦，复杂度q * SZ(x) * log(SZ(y))

对于SZ(x) > sqrt(n) 这样的 x ，个数绝对不超过sqrt(n)所以如果两两之间的答案全部算出来的

最坏复杂度是sqrt(n) * sqrt(n) / 2 * sqrt(n) * log(n) == n * sqrt(n) * log(n) 。所以直接记忆化就好啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

const int B = 330;

map<PII, LL> Map;
int treecnt, belong[N];
vector<int> tree[N];
vector<LL> sum[N];

int n, m, q, maxdis[N], dia[N], son[N];
vector<int> G[N];
bool root[N];

void dfs(int u, int fa, int idx) {
    belong[u] = idx;
    son[idx]++;
    for(auto& v : G[u]) {
        if(v == fa) continue;
        dfs(v, u, idx);
        chkmax(maxdis[u], maxdis[v] + 1);
    }
}

void dfs2(int u, int fa, int maxup) {
    chkmax(maxdis[u], maxup + 1);
    int mx0 = maxup, mx1 = -inf;
    for(auto& v : G[u]) {
        if(v == fa) continue;
        if(maxdis[v] > mx0) mx1 = mx0, mx0 = maxdis[v];
        else if(maxdis[v] > mx1) mx1 = maxdis[v];
    }
    for(auto& v : G[u]) {
        if(v == fa) continue;
        if(maxdis[v] == mx0) dfs2(v, u, mx1 + 1);
        else dfs2(v, u, mx0 + 1);
    }
}

LL calc(int u, int v) {
    LL ans = 0;
    LL maxdia = max(dia[u], dia[v]);
    for(auto& d : tree[u]) {
        int p = upper_bound(ALL(tree[v]), maxdia - d - 1) - tree[v].begin();
        ans += p * maxdia;
        if(p < SZ(tree[v])) {
            ans += (SZ(tree[v]) - p) * (d + 1) + sum[v].back();
            if(p - 1 >= 0) ans -= sum[v][p - 1];
        }
    }
    return ans;
}

int main() {
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i = 1; i <= n; i++)
        if(!belong[i]) dfs(i, 0, ++treecnt), root[i] = true;
    for(int i = 1; i <= n; i++) if(root[i]) dfs2(i, 0, -1);
    for(int i = 1; i <= n; i++) {
        tree[belong[i]].push_back(maxdis[i]);
        chkmax(dia[belong[i]], maxdis[i]);
    }
    for(int i = 1; i <= treecnt; i++) {
        sort(ALL(tree[i]));
        sum[i].resize(SZ(tree[i]));
        sum[i][0] = tree[i][0];
        for(int j = 1; j < SZ(sum[i]); j++)
            sum[i][j] = sum[i][j - 1] + tree[i][j];
    }
    for(int i = 1; i <= q; i++) {
        int u, v; scanf("%d%d", &u, &v);
        u = belong[u]; v = belong[v];
        if(son[u] > son[v]) swap(u, v);
        if(u == v) {
            puts("-1");
        } else {
            if(Map.find(mk(u, v)) == Map.end()) Map[mk(u, v)] = calc(u, v);
            printf("%.12f\n", 1.0 * Map[mk(u, v)] / son[u] / son[v]);
        }
    }
    return 0;
}

/*
*/

Codeforces 840D Expected diameter of a tree 分块思想

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