玲珑杯”ACM比赛 Round #19 B 维护单调栈

Posted 2020-09-30 半根毛线code

1149 - Buildings

Time Limit：2s Memory Limit：128MByte

Submissions：588Solved：151

DESCRIPTION

There are $\mathrm{nn\; buildings\; lined\; up,\; and\; the\; height\; of\; the\; ii-th\; house\; is\; hihi.}$

An inteval $[l,r][l,r](l\le r)(l\le r)\; is\; harmonious\; if\; and\; only\; if\; max(hl,\dots ,hr)-min(hl,\dots ,hr)\le kmax(hl,\dots ,hr)-min(hl,\dots ,hr)\le k.$

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers $\mathrm{n(1\le n\le 2\times 105),k(0\le k\le 109)n(1\le n\le 2\times 105),k(0\le k\le 109).\; The\; second\; line\; contains\; nn\; integers\; hi(1\le hi\le 109)hi(1\le hi\le 109).}$

OUTPUT

Print a line of one number which means the answer.

SAMPLE INPUT

3 1

1 2 3

SAMPLE OUTPUT

5

HINT

Harmonious intervals are: $[1,1],[2,2],[3,3],[1,2],[2,3][1,1],[2,2],[3,3],[1,2],[2,3].$

题意：给你一个长度为n的序列 问有多少区间 使得 区间最大值-区间最小值<=k

题解：单调栈处理出以a[i]为最小值的区间左界右界  组合出合法的区间 注意 （同一左界右界)或者称为同一块 下的最小值可能会有重复

从左向右遍历时 将当前值的左界改为(同一块中上一个相同值的位置+1) 具体看代码;

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <cctype>
 9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <bitset>
13 #include <string>
14 #include <complex>
15 #define ll long long
16 #define mod 1000000007
17 using namespace std;
18 ll n,k;
19 ll a[200005];
20 ll l[200005],r[200005];
21 map<ll,map<ll,ll>> mp;
22 int main()
23 {
24         scanf("%lld %lld",&n,&k);
25         for(int i=1; i<=n; i++)
26             scanf("%lld",&a[i]);
27         a[0]=-1ll;
28         a[n+1]=-1ll;
29         l[1]=1ll;
30         for(int i=2; i<=n; i++) //关键********
31         {
32             ll temp=i-1;
33             while(a[temp]>=a[i])//维护一个递增的序列
34                 temp=l[temp]-1;
35             l[i]=temp+1;
36         }
37         r[n]=n;
38         for (int i=n-1; i>=1; i--)
39         {
40             ll temp=i+1;
41             while(a[temp]>=a[i])
42                 temp=r[temp]+1;
43             r[i]=temp-1;
44         }
45         ll ans=0;
46         for(int i=1; i<=n; i++)
47         {
48             ll x=0,y=0;
49             if(mp[l[i]][r[i]]!=0){//去重 更改l[i]
50               ll now=l[i];
51               l[i]=mp[l[i]][r[i]];
52               mp[now][r[i]]=i+1;
53              }
54              else
55             mp[l[i]][r[i]]=i+1;
56             for(int j=i-1; j>=l[i]; j--)
57             {
58                 if((a[j]-a[i])<=k)
59                     x++;
60                 else
61                     break;
62             }
63             for(int j=i+1; j<=r[i]; j++)
64             {
65                 if((a[j]-a[i])<=k)
66                     y++;
67                 else
68                     break;
69             }
70             ans=ans+x*y+x+y+1ll;
71         }
72         printf("%lld\n",ans);
73     return 0;
74 }