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 << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, q;
int son
[2], w
;
vector<pair<int, int> > Q
;

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 right

vector<int> xs;
pair<int, int> ans
;

//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;
}