P2422 良好的感受

P2422 良好的感受

給定一段序列， 其中元素 \(0 \leq a_{i} \leq 100000\)
定義一段子段 \([L, R]\) 的溫馨值爲 \(\min_{L \leq i \leq R}a_{i} * \sum_{i = L}^{R}a_{i}\)
求最大溫馨值（並定位其位置）
\(n \leq 500000\)

Solution

首先 \(O(n^{2})\) 枚舉區間， 前綴和亂搞啥的顯然不合理
考慮式子， 後一項咱們能夠前綴和求出， 因而重點放在 \(\min_{L \leq i \leq R}a_{i}\) 上
對於一個 \(a_{i}\)， 貪心可知覆蓋越廣溫馨值越大
枚舉每個 \(a_{i}\) 做爲區間最小值， 而後能夠單調棧 \(O(1)\) 求出每一個 \(d_{i}\) 對應的最長左區間
同理能夠 \(O(n)\) 求出對應右端點
總複雜度 \(O(n)\)

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 200019, INF = 0xfffffffffffffff;
LL num;
LL a[maxn], sum[maxn];
LL l[maxn], r[maxn];
struct Stack{
    LL Index, val;
    }S[maxn];
LL top;
void get_l(){
    top = 0;
    S[++top].Index = 0;
    S[top].val = -INF;
    REP(i, 1, num){
        while(top >= 1 && S[top].val >= a[i])top--;
        l[i] = S[top].Index + 1;
        S[++top].Index = i;
        S[top].val = a[i];
        }
    }
void get_r(){
    top = 0;
    S[++top].Index = num + 1;
    S[top].val = -INF;
    for(LL i = num;i >= 1;i--){
        while(top >= 1 && S[top].val >= a[i])top--;
        r[i] = S[top].Index - 1;
        S[++top].Index = i;
        S[top].val = a[i];
        }
    }
int main(){
    while(scanf("%lld", &num) != EOF){
        memset(sum, 0, sizeof(sum));
        memset(a, 0, sizeof(a));
        memset(l, 0, sizeof(l));
        memset(r, 0, sizeof(r));
        REP(i, 1, num)a[i] = RD(), sum[i] = sum[i - 1] + a[i];
        get_l();
        get_r();
        LL ans = 0, L = 0, R = 0;
        REP(i, 1, num){
            if((sum[r[i]] - sum[l[i] - 1]) * a[i] >= ans){
                ans = (sum[r[i]] - sum[l[i] - 1]) * a[i];
                L = l[i], R = r[i];
                }
            }
        printf("%lld\n%lld %lld\n\n", ans, L, R);       
        }
    return 0;
    }