2019ACM-ICPC徐州网络赛-E-XKC's basketball team（set+贪心）

 E-XKC's basketball team

XKC , the captain of the basketball team , is directing a train of nn team members. He makes all members stand in a row , and numbers them 1 \cdots n1⋯n from left to right.

The ability of the ii-th person is w_iwi​ , and if there is a guy whose ability is not less than w_i+mwi​+mstands on his right , he will become angry. It means that the jj-th person will make the ii-th person angry if j>ij>i and w_j \ge w_i+mwj​≥wi​+m.

We define the anger of the ii-th person as the number of people between him and the person , who makes him angry and the distance from him is the longest in those people. If there is no one who makes him angry , his anger is -1−1 .

Please calculate the anger of every team member .

Input

The first line contains two integers nn and m(2\leq n\leq 5*10^5, 0\leq m \leq 10^9)m(2≤n≤5∗105,0≤m≤109) .

The following  line contain nn integers w_1..w_n(0\leq w_i \leq 10^9)w1​..wn​(0≤wi​≤109) .

Output

A row of nn integers separated by spaces , representing the anger of every member .

样例输入复制

6 1
3 4 5 6 2 10

样例输出复制

4 3 2 1 0 -1

思路

标程用的是单调队列+二分，我用的是set+map，想法差不多。从后往前遍历arr[],严格单调递增插入arr[i],并用map将arr[i]的下标存起来,
ans[i]=lower_bound(arr[i]+m)==set.end()?-1:lower_bound(arr[i]+m)

 1 #include 
 2 #pragma GCC optimize(3)
 3 using namespace std;
 4 typedef long long ll;
 5 const int maxn=1e6+7;
 6 set<int> ste;
 7 typedef set<int>::iterator sii;
 8 map<int,int> MAP;
 9 int ans[maxn];
10 int arr[maxn];
11 int main()
12 {
13     int n,m;
14     scanf("%d%d",&n,&m);
15     for(int i=1;i<=n;++i)
16     {
17         scanf("%d",&arr[i]);
18     }
19     for(int i=n;i>0;--i)
20     {
21         sii it=ste.lower_bound(arr[i]+m);
22         if(it!=ste.end())
23         ans[i]=MAP[*it]-i-1;
24         else ans[i]=-1;
25         if(ste.empty())
26         {
27             ste.insert(arr[i]);
28             MAP[arr[i]]=i;
29         }
30         else
31         {
32             it=--ste.end();
33             if(arr[i]>(*it))
34             {
35                 ste.insert(arr[i]);
36                 MAP[arr[i]]=i;
37             }
38         }
39     }
40     for(int i=1;i<=n;++i)
41     {
42         printf("%d",ans[i]);
43         if(i" ");
44         else printf("\n");
45     }
46     return 0;
47 }
48 /*
49 5 2
50 5 4 3 2 1
51 */

 

转载于:https://www.cnblogs.com/CharlieWade/p/11482359.html