hdu 6058 Kanade's sum

Kanade's sum

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 516    Accepted Submission(s): 182

Problem Description

Give you an array A[1..n]of length n. 

Let f(l,r,k) be the k-th largest element of A[l..r].

Specially , f(l,r,k)=0 if r−l+1<k.

Give you k , you need to calculate ∑nl=1∑nr=lf(l,r,k)

There are T test cases.

1≤T≤10

k≤min(n,80)

A[1..n] is a permutation of [1..n]

∑n≤5∗105

 

Input

There is only one integer T on first line.

For each test case,there are only two integers n,k on first line,and the second line consists of n integers which means the array A[1..n]

 

Output

For each test case,output an integer, which means the answer.

 

Sample Input

15 21 2 3 4 5

 

Sample Output

30

 

Source

2017 Multi-University Training Contest - Team 3

 

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Statistic | Submit | Discuss | Note题意：求所有区间的第k大的权值和，不存在第k大权值为0。

思路：问题等价于求每个数字分别是多少个区间的第k大，也就是可以枚举位置，然后再枚举左边x个数比这个数要大，那么右边必须要有k-1-x个数比这个位置要大，思路大概是这样。因为序列是1~n的排列，我们先记录每个数字出现的位置，我们维护一个链表，从最大值n开始枚举到1，每次把值插入到链表中，然后通过遍历链表用滑动窗口的办法达到O（n*（logn+k）），那个logn是我用set来记录已经插入的坐标有哪些，然后可以在logn的时间内找到第一个比当前插入坐标要大的坐标，从而达到修改链表的操作。

具体方法看代码吧。也有标注释，不懂留言。

//#pragma comment(linker, "/STACK:102400000,102400000") #include<iostream>  #include<cmath>  #include<queue>  #include<cstdio>  #include<queue>  #include<algorithm>  #include<cstring>  #include<string>  #include<utility>#include<set>#include<map>#include<stack>#include<vector>#define maxn 500005#define inf 0x3f3f3f3fusing namespace std;typedef long long LL;const double eps = 1e-8;const int mod = 1e9 + 7;int n, left1[maxn], right1[maxn], vis[maxn];const int read(){char ch = getchar();while (ch<'0' || ch>'9') ch = getchar();int x = ch - '0';while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';return x;}int main(){int t;scanf("%d", &t);while (t--){LL ans = 0;int k;scanf("%d%d", &n, &k);set<int>s;set<int>::iterator it;for (int i = 1; i <= n; i++){int x = read();left1[i] = 0;right1[i] = n + 1;vis[x] = i;//标记每个数字所在的位置}if (k < 1)printf("0\n");left1[n + 1] = 0;right1[n + 1] = n + 1;s.insert(0);s.insert(n + 1);//默认两位第0位和n+1位最大for (int i = n; i > 0; i--){s.insert(vis[i]);it = s.find(vis[i]);int l , r;it++;//找到第一个比当前位置大的位置r = *it;l = left1[r];left1[r] = vis[i];right1[vis[i]] = r;right1[l] = vis[i];left1[vis[i]] = l;//链表的插入int nowl = vis[i];for (int j = 0; j < k&&nowl; j++)nowl = left1[nowl];int nowr = nowl;for (int j = 0; j < k&&nowr != n + 1; j++)nowr = right1[nowr];//找到左端点和右端点for (int j = 0; j < k; j++){if (nowl == vis[i] || nowr == n + 1)break;int nextl = right1[nowl];int nextr = right1[nowr];ans += 1ll*(nextl - nowl)*(nextr - nowr)*i;nowl = nextl;nowr = nextr;}}printf("%lld\n", ans);}}