cf1359D. Yet Another Yet Another Task

链接

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题解

就处理下每个数在哪个区间里能作为最大值

然后求“强制包含第 $i$个数的和最大的区间”，且左端点和右端点的位置都有限制

以上所有操作都用 $ST$表就能解决

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct ST
{
    vector<vector<ll>> st;
    ll (*cmp)(ll,ll);
    ST(vector<ll> v, ll (*Cmp)(ll,ll)=[](ll a, ll b){return min(a,b);})
    {
        ll i, k, N=v.size();
        cmp=Cmp;
        st.emb(v);
        for(k=1;(1<<k)<=N;k++)
        {
            st.emb( vector<ll>(N-(1<<k)+1) );
            rep(i,0,N-(1<<k))
            {
                st[k][i]=cmp(st[k-1][i],st[k-1][i+(1<<k-1)]);
            }
        }
    }
    ll q(ll l, ll r)
    {
        if(l>r)swap(l,r);
        ll t=floor(log2(r-l+1));
        return cmp(st[t][l],st[t][r-(1<<t)+1]);
    }
};
int main()
{
    ll n=read(), i, L, R;
    vector<ll> a(n+10), s(n+10), l(n+10), r(n+10), pre(n+10), suf(n+10);
    rep(i,1,n)a[i]=read();
    rep(i,1,n)pre[i]=pre[i-1]+a[i];
    drep(i,n,1)suf[i]=suf[i+1]+a[i];
    ST stpre(pre), stsuf(suf), sta(a,[](ll x,ll y){return max(x,y);});
    rep(i,1,n)
    {
        L=1, R=i;
        while(L<R)
        {
            ll mid(L+R>>1);
            if(sta.q(mid,i)==a[i])R=mid;
            else L=mid+1;
        }
        l[i]=L;
        
        L=i, R=n;
        while(L<R)
        {
            ll mid(L+R+1>>1);
            if(sta.q(i,mid)==a[i])L=mid;
            else R=mid-1;
        }
        r[i]=L;
    }
    ll ans=-linf;
    rep(i,1,n)
    {
        ll t = pre[i-1] - stpre.q(l[i]-1,i-1) + suf[i+1] - stsuf.q(i+1,r[i]+1);
        ans = max(ans,t);
    }
    printf("%lld",ans);
    return 0;
}