HDU 5654 xiaoxin and his watermelon candy(离线思想、BIT)

题意:

$N,Q\le 2\times 10^5,N个数的序列,A_i\in[0,10^9]$
$定义奇怪的三元组为(i,j,k),i,j,k连续,且A_i\le A_j\le A_k$
$询问区间[l,r]中不同的奇怪三元组的个数$

分析:

$离线sb套路题,询问右端点排序$
$从左往右扫描A_i,一旦找到合法的三元组$
$BIT维护出现的个数,更新要在三元组的第1个位置$
$就把之前的位置-1,现在的位置+1$
$现在所有以i为右端点的询问的答案,只要查询sum(L,i)就可以了$
$即ans(L,i)=sum(L,i),这里BIT倒过来用就可以了,向前更新,向后查询$
$当然你要主席树强制在线也是兹磁的,反正复杂度一样的$
$时间复杂度为O(nlogn)$

代码:

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//
// Created by TaoSama on 2016-04-07
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#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 = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, q, a[N];
struct BIT {
int n, b[N];
void init(int _n) {
n = _n;
memset(b, 0, sizeof b);
}
void add(int i, int v) {
for(; i; i -= i & -i) b[i] += v;
}
int sum(int i) {
int ret = 0;
for(; i <= n; i += i & -i) ret += b[i];
return ret;
}
} bit;
int ans[N];
vector<pair<int, int> > qs[N];
typedef tuple<int, int, int> tripe;
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);
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
scanf("%d", &q);
for(int i = 1; i <= n; ++i) qs[i].clear();
for(int i = 1; i <= q; ++i) {
int l, r; scanf("%d%d", &l, &r);
qs[r].push_back({l, i});
}
bit.init(n);
int cnt = 0;
map<tripe, int> mp;
for(int i = 1; i <= n; ++i) {
cnt = a[i] >= a[i - 1] ? cnt + 1 : 1;
if(cnt > 2) {
tripe cur = tripe(a[i - 2], a[i - 1], a[i]);
if(mp.count(cur)) bit.add(mp[cur], -1);
mp[cur] = i - 2;
bit.add(i - 2, 1);
}
for(auto& q : qs[i]) ans[q.second] = bit.sum(q.first);
}
for(int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}
return 0;
}


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