HDU 4605 Magic Ball Game（离线、BIT）

题意:

根为1的N≤105个节点的无向树,所有结点有2个儿子或者没有儿子
每个节点的重量wi≤109,然后有一个球,从根开始往儿子结点走
每碰到一个节点，有三种情况:
如果此球重量等于该节点重量或者没有儿子节点了，球就停下了
如果此球重量小于该节点重量，则分别往左右儿子走的可能都是1/2
如果此球重量大于该节点重量，则走向左儿子的概率是1/8，右儿子的概率是7/8
Q≤105询问,问一个重量为x的球经过节点v的概率,表示成7x/2y,输出x,y

分析:

首先树上2点之间的路径是唯一的,所以离线询问按照dfs序来回答询问
离散化之后用2颗BIT来记录左路径和右路径上经过的点的重量
ans=12leftLarge∗18leftSmall∗12rightLarge∗78rightSmall
即x=rightSmall,y=leftLarge+rightLarge+3∗(leftSmall+rightSmall)
递归之前添加,别忘记回溯就好
时间复杂度为O((n+q)logn)

代码:

////  Created by TaoSama on 2016-02-26//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, q;int son[N][2], w[N];vector<pair<int, int> > Q[N];struct BIT {    int n, b[N << 1];    void init(int _n) {        n = _n;        memset(b, 0, sizeof b);    }    void add(int i, int v) {        for(; i <= n; i += i & -i)            b[i] += v;    }    int sum(int i) {        int ret = 0;        for(; i; i -= i & -i) ret += b[i];        return ret;    }    int getAll() {        return sum(n);    }} bit[2];  //0 for left, 1 for rightvector<int> xs;pair<int, int> ans[N];//ans = 1/2 ^ leftLarge  * 1/8 ^ leftSmall;//    * 1/2 ^ rightLarge * 7/8 ^ rightSmall;void dfs(int u) {    int leftAll = bit[0].getAll(), rightAll = bit[1].getAll();    for(auto p : Q[u]) { //answer queries        int x = p.first, id = p.second;        x = lower_bound(xs.begin(), xs.end(), x) - xs.begin() + 1;        int leftSmall = bit[0].sum(x - 1), leftLarge = leftAll - bit[0].sum(x);        int rightSmall = bit[1].sum(x - 1), rightLarge = rightAll - bit[1].sum(x);        //if equal ones exist        if(leftAll + rightAll - leftSmall - leftLarge - rightSmall - rightLarge) {            ans[id] = { -1, -1};            continue;        }        ans[id].first = rightSmall;        ans[id].second = leftLarge + rightLarge + 3 * (leftSmall + rightSmall);    }    w[u] = lower_bound(xs.begin(), xs.end(), w[u]) - xs.begin() + 1;    for(int i = 0; i < 2; ++i) {        int v = son[u][i];        if(!v) continue;        bit[i].add(w[u], 1);  //add before go down        dfs(v);        bit[i].add(w[u], -1); //back    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d", &n);        xs.clear();        for(int i = 1; i <= n; ++i) {            scanf("%d", w + i);            xs.push_back(w[i]);        }        scanf("%d", &m);        memset(son, 0, sizeof son);        while(m--) {            int u; scanf("%d", &u);            for(int i = 0; i < 2; ++i)                scanf("%d", son[u] + i);        }        scanf("%d", &q);        for(int i = 1; i <= n; ++i) Q[i].clear();        for(int i = 1; i <= q; ++i) {            int v, x; scanf("%d%d", &v, &x);            xs.push_back(x);            Q[v].push_back({x, i});        }        sort(xs.begin(), xs.end());        xs.resize(unique(xs.begin(), xs.end()) - xs.begin());        for(int i = 0; i < 2; ++i) bit[i].init(xs.size());        dfs(1);        for(int i = 1; i <= q; ++i)            if(~ans[i].first) printf("%d %d\n", ans[i].first, ans[i].second);            else puts("0");    }    return 0;}