HDU-2017 多校训练赛9-1002-Ch’s gift

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ACM模版

描述

题解

这个题做法五花八门的，有的人用树链剖分+扫描线过的，有的大佬是用树状数组过的，反正方法太多了，我也很懵逼……而我，用的是 LCA+主席树搞得，这个解法真是个野路子，虽然 AC 了，但是花了我最后的两个多小时调试……

代码

#include <iostream>#include <vector>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 1e5 + 7;int A[MAXN];int B[MAXN];int pre[MAXN];int id[MAXN];int d[MAXN];int ls[MAXN * 20];int rs[MAXN * 20];int T[MAXN * 20];int anc[MAXN][25];ll cnt[MAXN * 20];vector<int> vi[MAXN];int ql, qr, tol, tot;void init(int n){    tot = 1;    tol = 0;    for (int i = 1; i <= n; i++)    {        vi[i].clear();    }}int low(int *a, int pos, int R){    int l = 1, r = R;    while (l <= r)    {        int mid = (l + r) >> 1;        if (a[mid] == pos)        {            return mid;        }        else if (a[mid] > pos)        {            r = mid - 1;        }        else        {            l = mid + 1;        }    }    return 0;}void builde(int l, int r, int &o){    o = tol++;    if (l == r)    {        cnt[o] = 0;        return ;    }    int m = (l + r) >> 1;    builde(l, m, ls[o]);    builde(m + 1, r, rs[o]);    cnt[o] = 0;}void update(int last, int l, int r, int pos, int &o){    o = tol++;    cnt[o] = cnt[last] + A[pos];    ls[o] = ls[last];    rs[o] = rs[last];    if (l == r)    {        return ;    }    int m = (l + r) >> 1;    if (m >= pos)    {        update(ls[last], l, m, pos, ls[o]);    }    else    {        update(rs[last], m + 1, r, pos, rs[o]);    }}ll query(int ll, int rr, int l, int r){    if (ql <= l && qr >= r)    {        return cnt[rr] - cnt[ll];    }    int m = (l + r) >> 1;    long long ans = 0;    if (ql <= m)    {        ans += query(ls[ll], ls[rr], l, m);    }    if (qr > m)    {        ans += query(rs[ll], rs[rr], m + 1, r);    }    return ans;}int LCA(int p, int q){    if (d[p] < d[q])    {        swap(p, q);    }    int log;    for (log = 1; (1 << log) <= d[p]; log++) ;    log--;    for (int i = log; i >= 0; i--)    {        if (d[p] - (1 << i) >= d[q])        {            p = anc[p][i];        }    }    if (p == q)    {        return p;    }    for (int i = log; i >= 0; i--)    {        if (anc[p][i] != -1 && anc[p][i] != anc[q][i])        {            p = anc[p][i];            q = anc[q][i];        }    }    return anc[p][0];}void preprocess(int n){    for (int i = 1; i <= n; i++)    {        anc[i][0] = pre[i];        for (int j = 1; (1 << j) < n; j++)        {            anc[i][j] = -1;        }    }    for (int j = 1; (1 << j) <= n; j++)    {        for (int i = 1; i <= n; i++)        {            if (anc[i][j - 1] != -1)            {                int a = anc[i][j - 1];                anc[i][j] = anc[a][j - 1];            }        }    }}void dfs(int u, int p, int m){    d[u] = d[p] + 1;    pre[u] = p;    id[u] = tot++;    int pos = low(A, B[u], m);    update(T[id[p]], 1, m, pos, T[id[u]]);    for (int i = 0; i < vi[u].size(); i++)    {        int v = vi[u][i];        if (v == p)        {            continue;        }        dfs(v, u, m);    }}int low1(int *a, int pos, int R){    int l = 1, r = R;    while (l < r)    {        int m = (l + r) >> 1;        if (a[m] >= pos)        {            r = m;        }        else        {            l = m + 1;        }    }    return l;}int n, q;int main(){    while (~scanf("%d%d", &n, &q))    {        init(n);        for (int i = 1; i <= n; i++)        {            scanf("%d", &A[i]);            B[i] = A[i];        }        sort(A + 1, A + 1 + n);        int m = 1;        for (int i = 2; i <= n; i++)        {            if (A[i] != A[i - 1])            {                A[++m] = A[i];            }        }        int u, v;        for (int i = 1; i < n; i++)        {            scanf("%d%d", &u, &v);            vi[u].push_back(v);            vi[v].push_back(u);        }        builde(1, m, T[0]);        d[0] = 0;        dfs(1, 0, m);        preprocess(n);        ll ans;        int s, t, a, b;        for (int i = 1; i <= q; i++)        {            scanf("%d%d%d%d", &s, &t, &a, &b);            ql = low1(A, a, m);            qr = low1(A, b, m);            if (A[qr] > b)            {                qr--;            }            int o = LCA(s, t);            ans = 0;            ans += query(T[id[pre[o]]], T[id[s]], 1, m);            ans += query(T[id[o]], T[id[t]], 1, m);            if (i == 1)            {                printf("%lld", ans);            }            else            {                printf(" %lld", ans);            }        }        putchar(10);    }    return 0;}

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