HDU6058 Kanade's sum（链表）-优快云博客

Kanade's sum

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2859 Accepted Submission(s): 1182

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

Sample Output

Source

2017 Multi-University Training Contest - Team 3

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liuyiding

题意：给一个数组，找到所有区间的第k大数求和

将问题转化一下，可以变为某一个数可以在多少个区间成为第k大数。

对于每一个数x，我们只需要在x的左边找k个比x大的数，x的右边找k个比x大的数，线性扫一遍就能知道有多少个区间包含x而且x是区间第k大。

用链表维护，从小到大枚举所以可以直接删除。

#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int MAXN=1000100;
const int mod=1e9+7;
int a[MAXN],pos[MAXN];
int pre[MAXN],nxt[MAXN];
void del(int x){
    pre[nxt[x]]=pre[x];
    nxt[pre[x]]=nxt[x];
}

int n,k;
long long L[100],R[100];
long long solve(int x){
    int l=0,r=0;
    //往前跳k步
    for(int i=x;i&&l<=k;i=pre[i]){
        L[++l]=i-pre[i];//中间距离多少个数
    }
    //往后跳k步
    for(int i=x;i<=n&&r<=k;i=nxt[i]){
        R[++r]=nxt[i]-i;
    }
    long long ans=0;
    for(int i=1;i<=l;i++){
        if(k-i+1<=r&&k-i+1>=1){
            ans+=L[i]*R[k-i+1];
        }
    }
    return ans;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
         scanf("%d%d",&n,&k);
         for(int i=1;i<=n;i++){
            scanf("%d",a+i);
            pos[a[i]]=i;
         }
         for(int i=0;i<=n+1;i++){
            pre[i]=i-1;
            nxt[i]=i+1;
         }
         pre[0]=0,nxt[n+1]=n+1;
         long long res=0;
         for(int i=1;i<=n;i++){
            res+=solve(pos[i])*i;
            //因为是从小到大来算的,所以可以直接删除
            del(pos[i]);
         }
         printf("%lld\n",res);
    }
}