HDU 5289 Assignment(two pointers)

题意:

$N\le 10^5的序列,A_i\le 10^9$
$求连续区间中任意2个数差值不超过k的区间个数$

分析:

$two pointers经典题辣,还是这种写法$
$对于每个左端点,找到最右的右端点统计贡献数即可$
$时间复杂度O(n)$

代码:

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//
// Created by TaoSama on 2016-04-12
// 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 = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, a[N], k;
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%d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
multiset<int> s;
long long ans = 0;
for(int l = 1, r = 1; l <= n; ++l) {
while(r <= n) {
if(s.size()) {
if(a[r] - *s.begin() >= k || *s.rbegin() - a[r] >= k)
break;
}
s.insert(a[r++]);
}
ans += r - l;
s.erase(s.find(a[l]));
}
printf("%I64d\n", ans);
}
return 0;
}


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