P2422 良好的感觉

P2422 良好的感觉

给定一段序列， 其中元素 \(0 \leq a_{i} \leq 100000\)

定义一段子段 \([L, R]\) 的舒适值为 \(\min_{L \leq i \leq R}a_{i} * \sum_{i = L}^{R}a_{i}\)

求最大舒适值（并定位其位置）

\(n \leq 500000\)

Solution

首先 \(O(n^{2})\) 枚举区间， 前缀和乱搞啥的显然不合理

考虑式子， 后一项我们可以前缀和求出， 于是重点放在 \(\min_{L \leq i \leq R}a_{i}\) 上

对于一个 \(a_{i}\)， 贪心可知覆盖越广舒适值越大

枚举每一个 \(a_{i}\) 作为区间最小值， 然后可以单调栈 \(O(1)\) 求出每个 \(d_{i}\) 对应的最长左区间

同理可以 \(O(n)\) 求出对应右端点

总复杂度 \(O(n)\)

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;
	}