2019南昌邀請賽網絡賽 I.Max answer

單調棧維護左邊和右邊第一個大於它的位置
用ST表維護區間前綴和的最大值和最小值
當前點小於0的時候就是$後半部分Smin-前半部分Smax$
大於0的時候就是$後半部分Smax-前半部分Smin$

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 5e5+1000;

ll stm[maxn][23];
ll stn[maxn][23];

int n;
int a[maxn];
ll s[maxn];
int l[maxn],r[maxn];


ll Querym(int l,int r)
{
    int k=log2(r-l+1); 
    return max(stm[l][k],stm[r-(1<<k)+1][k]);
}

ll Queryn(int l,int r)
{
    int k=log2(r-l+1); 
    return min(stn[l][k],stn[r-(1<<k)+1][k]);
}


int main()
{
    cin >> n;
    rep(i,1,n+1) cin >> a[i];
    rep(i,1,n+1) s[i] = s[i-1] + a[i];
    rep(i,0,n+1) stm[i][0] = stn[i][0] = s[i];
    for(int j=1;j<=21;j++)
        for(int i=0;i+(1<<j)-1<=n;i++)
            {
            	stm[i][j]=max(stm[i][j-1],stm[i+(1<<(j-1))][j-1]);
            	stn[i][j]=min(stn[i][j-1],stn[i+(1<<(j-1))][j-1]);
            }
    stack<int> sta;

   	rep(i,1,n+1)
    {
        while (!sta.empty()&&a[i]<=a[sta.top()]) sta.pop();
        if (sta.empty()) l[i]=1; else l[i]=sta.top()+1;
        sta.push(i);
    }
    while (!sta.empty()) sta.pop();
   	per(i,1,n+1)
    {
        while (!sta.empty()&&a[i]<=a[sta.top()]) sta.pop();
        if (sta.empty()) r[i]=n; else r[i]=sta.top()-1;
        sta.push(i);
    }
    ll ans = -1e16;
    rep(i,1,n+1)
    {
    	if(a[i] > 0)
    	{
    		ll rrr = Querym(i,r[i]);
    		ll lll = Queryn(l[i]-1,i-1);
    		ans = max(ans,(rrr-lll)*(ll)a[i]);
    	}
    	else if(a[i] < 0)
    	{
    		ll rrr = Queryn(i,r[i]);
    		ll lll = Querym(l[i]-1,i-1);
    		ans = max(ans,(rrr-lll)*(ll)a[i]);	
    	}
    	else
    	{
    		ans = max(0ll,ans);
    	}
    }
    cout << ans << endl;
}