HDU 4605 (主席树)

题目链接：点击这里

题意：有一棵树二叉，每个节点有一个数字。每次询问一个节点和一个数字，问这个数字经过这个节点的概率。一个数字从根开始往下走，如果走过的节点数字和他相同，就停在这里；如果节点数字比他大，向左向右的概率都是1/2；否则向左的概率是1/8，向右的概率是7/8。

对于每个询问，需要统计他到根节点路径之间的信息。所以从根开始开主席数，线段数记录走向左儿子的某个数字的个数和走向右儿子的某个数字的个数。这样对于每一个询问，找到a,b,c,d四个值，分别代表向左走的比询问大的数，向右走的比询问大的数，向左走的比询问小的数，向右走的比询问大的数。无解当且仅当询问到根路径有等于询问的数；否则把这四个数整理一下就是答案了。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <queue>using namespace std;#define N 100005#define maxn 200005#define maxm maxn*60struct node {    int l, r;    int lnum, rnum;}tree[maxm];int tot, T[N];int n, m, q;int Left[N], Right[N], fa[N];int qu[N][2];vector <int> num;int gg[maxn], cnt, w[N];void lisanhua () {///离散化    sort (num.begin(), num.end ());    int sz = num.size ();    cnt = 0;    for (int i = 0; i < sz; i++) {        if (!i || num[i] != num[i-1]) {            gg[++cnt] = num[i];        }    }    for (int i = 1; i <= n; i++) {        w[i] = lower_bound (gg+1, gg+1+cnt, w[i])-gg;    }    for (int i = 1; i <= q; i++) {        qu[i][1] = lower_bound (gg+1, gg+1+cnt, qu[i][1])-gg;    }}int build_tree (int l, int r) {    int root = tot++;    tree[root].lnum = tree[root].rnum = 0;    if (l == r) return root;    int mid = (l+r)>>1;    tree[root].l = build_tree (l, mid);    tree[root].r = build_tree (mid+1, r);    return root;}int update (int root, int pos, int val, int from) {    int new_root = tot++, tmp = new_root;    int l = 1, r = cnt;    tree[new_root].lnum = tree[root].lnum;    tree[new_root].rnum = tree[root].rnum;    if (from == 0) tree[new_root].lnum += val;    else tree[new_root].rnum += val;    while (l < r) {        int mid = (l+r)>>1;        if (mid >= pos) {            tree[new_root].r = tree[root].r;            tree[new_root].l = tot++;            root = tree[root].l;            new_root = tree[new_root].l;            r = mid;        }        else {            tree[new_root].l = tree[root].l;            tree[new_root].r = tot++;            root = tree[root].r;            new_root = tree[new_root].r;            l = mid+1;        }        tree[new_root].lnum = tree[root].lnum;        tree[new_root].rnum = tree[root].rnum;        if (from == 0) {            tree[new_root].lnum += val;        }        else tree[new_root].rnum += val;    }    return tmp;}int d[N];struct Q {    int u, from;};void bfs (int s) {     d[s] = 0;    T[s] = T[n+1];    queue <Q> q; while (!q.empty ()) q.pop (); q.push ((Q){s, 0});    while (!q.empty ()) {        Q tmp = q.front (); q.pop (); int u = tmp.u; int from = tmp.from;        if (fa[u] != -1) T[u] = update (T[fa[u]], w[fa[u]], 1, from);        if (Left[u] != -1) {            d[Left[u]] = d[u]+1;            q.push ((Q){Left[u], 0});        }        if (Right[u] != -1) {            d[Right[u]] = d[u]+1;            q.push ((Q){Right[u], 1});        }    }}int query_more (int root, int num, int from) {    int ans = 0;    int l = 1, r = cnt;    while (r >= l) {        if (l == r) {            if (l > num) {                if (from == 0) ans += tree[root].lnum;                else ans += tree[root].rnum;            }            break;        }        int mid = (l+r)>>1;        if (mid <= num) {            root = tree[root].r;            l = mid+1;        }        else {            if (from == 0) ans += tree[tree[root].r].lnum;            else ans += tree[tree[root].r].rnum;            root = tree[root].l;            r = mid;        }    }    return ans;}int query_less (int root, int num, int from) {    int ans = 0;    int l = 1, r = cnt, mid = (l+r)>>1;    while (r >= l) {        if (l == r) {            if (l < num) {                if (from == 0) ans += tree[root].lnum;                else ans += tree[root].rnum;            }            break;        }        int mid = (l+r)>>1;        if (mid >= num) {            root = tree[root].l;            r = mid;        }        else {            if (from == 0) ans += tree[tree[root].l].lnum;            else ans += tree[tree[root].l].rnum;            root = tree[root].r;            l = mid+1;        }    }    return ans;}void solve () {    tot = 0;    T[n+1] = build_tree (1, cnt);    int root;    for (int i = 1; i <= n; i++) if (fa[i] == -1) {        root = i;        break;    }    bfs (root);    for (int i = 1; i <= q; i++) { //cout << i << endl;        int u = qu[i][0], num = qu[i][1], father = (u == root ? n+1 : fa[u]);        int aa, bb, cc, dd;        aa = query_more (T[u], num, 0), bb = query_more (T[u], num, 1);        cc = query_less (T[u], num, 0), dd = query_less (T[u], num, 1);        int equ = d[u]-aa-bb-cc-dd;//和这个数字相等的        if (equ) {            printf ("0\n");            continue;        }        printf ("%d %d\n", dd, aa+bb+3*(cc+dd));    }}int main () {    //freopen ("more.in", "r", stdin);    int t;    scanf ("%d", &t);    while (t--) {        scanf ("%d", &n);         num.clear ();        for (int i = 1; i <= n; i++) {            scanf ("%d", &w[i]);            num.push_back (w[i]);            Left[i] = Right[i] = fa[i] = -1;        }        scanf ("%d", &m);        for (int i = 0; i < m; i++) {            int u, a, b; scanf ("%d%d%d", &u, &a, &b);            Left[u] = a, Right[u] = b;            fa[a] = u, fa[b] = u;        }        scanf ("%d", &q);        for (int i = 1; i <= q; i++) {            scanf ("%d%d", &qu[i][0], &qu[i][1]);            num.push_back (qu[i][1]);        }        lisanhua ();        solve ();    }    return 0;}