2021HNCPC-E-差分,思维

题目大意:

求解每一个$f(i)$代表$max(a_j,...,a_i)-(j-i+1)\ge m$的$j$的个数.$j\le i$
$n\le 1e6$

思路:

关键:看到$max(a_j,...,a_i)$,想到枚举$a_i$,考虑它所管辖的范围(单调栈求解).

PS：当有相同的值时，可以规定将区间贡献记在最右边的值上。

那么就转化成$(j-i+1)\le max(a_j,...,a_i)-m$.

然后枚举右半边，分情况讨论来区间加等差序列即可.

代码仓库:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define vll vector<ll>
#define fi first
#define se second
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
int a[maxn] , l[maxn] , r[maxn] , b[maxn];
int n , m;
template <typename T>
void read(T & x){
     x = 0;T f = 1;char ch = getchar();while(!isdigit(ch)){
    if(ch == '-') f = -1;
ch = getchar();}while (isdigit(ch)){
    x = x * 10 + (ch ^ 48);ch = getchar();}x *= f;}
template <typename T>
void write(T x){
    if(x < 0){
    putchar('-');x = -x;}if (x > 9)write(x / 10);putchar(x % 10 + '0');}
void add (int l , int r , int s , int d){
    
    if (l > r) return ;
    int t = s + (r - l) * d;
    b[l] += s;
    b[l+1] += d - s;
    b[r+1] -= d + t;
    b[r+2] += t;
}
void solve (){
    
    for (int i = 1 ; i <= n ; i++) b[i] = 0;
    for (int i = 1 ; i <= n ; i++){
    
        int g = a[i] - m;
        if (g <= 0) continue;
        int v = i - l[i] + 1;
        if (v >= g){
    
            add(i , min(i + g - 1 , r[i]) , g , -1);
        }else{
    
            int gap = g - v;
            int e1 = min(i + gap , r[i]);
            add(i , e1 , v , 0);
            add(e1 + 1 , min(r[i] , e1 + v - 1) , v - 1 , -1);
        }
    }
    for (int i = 1 ; i <= n ; i++) b[i] += b[i - 1];
    for (int i = 1 ; i <= n ; i++) b[i] += b[i - 1];
}
int main()
{
    
    while (~scanf("%d%d" , &n, &m)){
    
        for (int i = 1 ; i <= n ; i++) read(a[i]);
        stack<int> s;
        for (int i = 1 ; i <= n ; i++){
    
            while (s.size() && a[s.top()] <= a[i]) s.pop();
            if (s.size()) l[i] = s.top() + 1;
            else l[i] = 1;
            s.push(i);
        }
        while (s.size()) s.pop();
        for (int i = n ; i >= 1 ; i--){
    
            while (s.size() && a[i] > a[s.top()]) s.pop();
            if (s.size()) r[i] = s.top() - 1;
            else r[i] = n;
            s.push(i);
        }
        solve();
        for (int i = 1 ; i <= n ; i++) printf("%d " , b[i]);
        printf("\n");
    }
    return 0;
}