Codeforces 418D Big Problems for Organizers 树的直径 + st表（看题解）

Big Problems for Organizers

想了挺久没有想出来。。 感觉也是个套路， 以后树上两点最长路径要考虑一下把树的直径抠出来， 

会发现就可以分类讨论解决问题了。。

#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 = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
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;}

int n, m, pa[N], depth[N];
int dia[N], cntdia;
int mx[N], mxL[N], mxR[N];
int mxd[N];
int belong[N];
vector<int> G[N];
bool ban[N];

int Log[N];
struct ST {
    int dp[N][20]; int ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline int query(int x, int y) {
        if(x > y) return 0;
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
} rmqL, rmqR;

PII getDia(int u, int fa) {
    depth[u] = depth[fa] + 1;
    pa[u] = fa;
    PII ret = mk(depth[u], u);
    for(auto &v : G[u]) {
        if(v == fa) continue;
        chkmax(ret, getDia(v, u));
    }
    return ret;
}

void dfs(int u, int fa, int who) {
    belong[u] = who;
    depth[u] = depth[fa] + 1;
    chkmax(mx[who], depth[u]);
    for(auto &v : G[u]) {
        if(ban[v] || v == fa) continue;
        dfs(v, u, who);
        chkmax(mxd[u], mxd[v] + 1);
    }
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    int A = getDia(1, 0).se;
    int B = getDia(A, 0).se;
    while(B) dia[++cntdia] = B, B = pa[B];
    for(int i = 1; i <= cntdia; i++) {
        ban[dia[i]] = true;
        depth[dia[i]] = 0;
        belong[dia[i]] = i;
    }
    for(int i = 1; i <= cntdia; i++) {
        for(auto &v : G[dia[i]]) {
            if(ban[v]) continue;
            dfs(v, dia[i], i);
            chkmax(mxd[dia[i]], mxd[v] + 1);
        }
    }
    for(int i = 1; i <= cntdia; i++) {
        mxL[i] = mx[i] + (i - 1);
        mxR[i] = mx[i] + (cntdia - i);
    }
    rmqL.build(cntdia, mxL, 1);
    rmqR.build(cntdia, mxR, 1);
    scanf("%d", &m);
    while(m--) {
        int u, v, ans = 0;
        scanf("%d%d", &u, &v);
        int bu = belong[u];
        int bv = belong[v];
        if(bu == bv) {
            chkmax(ans, min(depth[u], depth[v]) + bu - 1);
            chkmax(ans, min(depth[u], depth[v]) + (cntdia - bu));
        } else {
            int tmpu = 0, tmpv = 0;
            if(bu > bv) {
                swap(bu, bv);
                swap(u, v);
            }
            int du = depth[u];
            int dv = depth[v];
            int dis = bv - bu;
            int mdis = (du + dv + dis) / 2;
            if(mdis < du) {
                chkmax(tmpv, rmqR.query(1, bv - 1) - (cntdia - bv));
                chkmax(tmpv, rmqL.query(bv + 1, cntdia) - (bv - 1));
            } else if(du + dis <= mdis) {
                chkmax(tmpu, rmqR.query(1, bu - 1) - (cntdia - bu));
                chkmax(tmpu, rmqL.query(bu + 1, cntdia) - (bu - 1));
            } else {
                int mid = bu + mdis - du;
                chkmax(tmpu, rmqR.query(1, bu - 1) - (cntdia - bu));
                chkmax(tmpu, rmqL.query(bu + 1, mid) - (bu - 1));
                chkmax(tmpv, rmqR.query(mid + 1, bv - 1) - (cntdia - bv));
                chkmax(tmpv, rmqL.query(bv + 1, cntdia) - (bv - 1));
            }
            chkmax(ans, tmpu + du);
            chkmax(ans, tmpv + dv);
            chkmax(ans, mxd[u]);
            chkmax(ans, mxd[v]);
        }
        printf("%d\n", ans);
    }
    return 0;
}

/*
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