南昌网络赛 Max answe 单调栈+线段树维护区间最大最小值

题目链接：https://nanti.jisuanke.com/t/38228

题意：求区间和*区间最小值 的最大值

题解：先单调栈求一下以每个位置为最小值的左右到哪，线段树维护下前缀的最大最小值，然后当前值为正值时，就让可选区间范围内区间的值最大，右边取个最大，左边取个最小，当前值为负数时，让可选区间的值最小，左边取个最小，右边取个最大

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+10;
typedef long long ll;
struct node
{
    int l,r;
    ll maxx,minn;
}tree[N<<2];
ll sum[N],a[N];
int n;
int l[N],r[N];
void pushup(int cur)
{
    tree[cur].maxx=max(tree[cur<<1].maxx,tree[cur<<1|1].maxx);
    tree[cur].minn=min(tree[cur<<1].minn,tree[cur<<1|1].minn);
}
void build(int l,int r,int cur)
{
    tree[cur].l=l;
    tree[cur].r=r;
    if(l==r)
    {
        tree[cur].maxx=tree[cur].minn=sum[l];
        return;
    }
    int mid=(r+l)>>1;
    build(l,mid,cur<<1);
    build(mid+1,r,cur<<1|1);
    pushup(cur);
 //   cout<<l<<" "<<r<<" "<<tree[cur].l<<" "<<tree[cur].maxx<<" "<<tree[cur].minn<<endl;
}
ll querymax(int pl,int pr,int cur)
{
    if(pl<=tree[cur].l&&tree[cur].r<=pr)
    {
        return tree[cur].maxx;
    }
    ll res=-1e18;
    if(pl<=tree[cur<<1].r) res=max(res,querymax(pl,pr,cur<<1));
    if(pr>=tree[cur<<1|1].l) res=max(res,querymax(pl,pr,cur<<1|1));
    return res;
}
ll querymin(int pl,int pr,int cur)
{
    if(pl<=tree[cur].l&&tree[cur].r<=pr)
    {
        return tree[cur].minn;
    }
    ll res=1e18;
    if(pl<=tree[cur<<1].r) res=min(res,querymin(pl,pr,cur<<1));
    if(pr>=tree[cur<<1|1].l) res=min(res,querymin(pl,pr,cur<<1|1));

    return res;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        stack<int> s;
        for(int i=1;i<=n;i++)
        {
            while(!s.empty() && a[s.top()] >= a[i]) s.pop();

            if(s.empty()) l[i]=0;
            else
            {
                l[i]=s.top();
           //    cout<<s.top()<<" "<<a[s.top()]<<" " <<a[i]<<" "<<(!s.empty() && a[s.top()] >= a[i])<<endl;
            }

            s.push(i);
       //     cout<<l[i]<<endl;
        }
        while(!s.empty()) s.pop();
        for(int i=n;i>=1;i--)
        {
            while(!s.empty() && a[s.top()] >= a[i]) s.pop();

            if(s.empty()) r[i]=n;
            else r[i]=s.top()-1;
         //   cout<<r[i]<<endl;
            s.push(i);
        }

        build(0,n,1);
        ll ans=-1e18,cnt;
        for(int i=1;i<=n;i++)
        {
            cnt=a[i];
         //   cout<<l[i]<<" "<<i-1<<" "<<i<<" "<<r[i]<<endl;
            if(cnt>0) cnt=cnt*(querymax(i,r[i],1)-querymin(l[i],i-1,1));
            else cnt=cnt*(querymin(i,r[i],1)-querymax(l[i],i-1,1));
        //    cout<<cnt<<endl;
            ans=max(ans,cnt);
        }
        printf("%lld\n",ans);
    }
    return 0;
}
/*
4
-3 -6 -9 1
4
2 -5 6 -9
*/