P2422 良好的感觉

Posted 2021-01-16 tony-double-sky

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
%lld %lld

", ans, L, R);       
        }
    return 0;
    }