nowcoder15162 小H的询问

这篇博客主要介绍了如何处理小H的一系列区间询问问题,包括维护强制包含左右端点的最优答案,整个区间的最优解,合法性判断以及区间和的计算。通过合并区间的方法来高效解决这些查询。

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链接

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题解

维护:强制包含左端点的最优答案,强制包含右端点的最有答案,整个区间的最优答案,整个区间是否合法,整个区间的和

合并两个区间:

Node merge(Node a, Node b)
    {
        if(a.l==-1)return b;
        if(b.l==-1)return a;
        Node c;
        c.l=a.l, c.r=b.r;
        c.sum = a.sum + b.sum;
        bool chigau = judge(arr[a.r],arr[b.l]);
        c.ok = a.ok and b.ok and chigau;
        c.lbest = chigau and a.ok ? max(a.sum+b.lbest,a.lbest) : a.lbest ;
        c.rbest = chigau and b.ok ? max(b.sum+a.rbest,b.rbest) : b.rbest ;
        c.best = max(a.best,b.best);
        if(chigau)c.best = max(c.best,a.rbest+b.lbest);
        return c;
    }

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 100010
#define maxe 100010
#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;
}
bool judge(ll a, ll b)
{
    return (a^b)&1;
}
ll arr[maxn];
struct SegmentTree
{
    struct Node
    {
        int l, r;
        ll sum, lbest, rbest, best;
        bool ok;
        Node(){l=-1;}
        Node(ll L, ll R, ll v)
        {
            l=L, r=R;
            sum=lbest=rbest=best=v;
            ok = true;
        }
    }node[maxn<<2];
    Node merge(Node a, Node b)
    {
        if(a.l==-1)return b;
        if(b.l==-1)return a;
        Node c;
        c.l=a.l, c.r=b.r;
        c.sum = a.sum + b.sum;
        bool chigau = judge(arr[a.r],arr[b.l]);
        c.ok = a.ok and b.ok and chigau;
        c.lbest = chigau and a.ok ? max(a.sum+b.lbest,a.lbest) : a.lbest ;
        c.rbest = chigau and b.ok ? max(b.sum+a.rbest,b.rbest) : b.rbest ;
        c.best = max(a.best,b.best);
        if(chigau)c.best = max(c.best,a.rbest+b.lbest);
        return c;
    }
    void pushup(ll o)
    {
        node[o] = merge( node[o<<1] , node[o<<1|1] );
    }
    void build(ll o, ll l, ll r)
    {
        ll mid(l+r>>1);
        node[o].l=l, node[o].r=r;
        if(l==r){node[o]=Node(l,l,arr[l]);return;}
        build(o<<1,l,mid);
        build(o<<1|1,mid+1,r);
        pushup(o);
    }
    Node Sum(ll o, ll l, ll r)
    {
        ll mid(node[o].l+node[o].r>>1);
        Node ans;
        if(l<=node[o].l and r>=node[o].r)return node[o];
        if(l<=mid)ans = merge(ans,Sum(o<<1,l,r));
        if(r>mid)ans = merge(ans,Sum(o<<1|1,l,r));
        return ans;
    }
    void chg(ll o, ll pos)
    {
        ll mid(node[o].l+node[o].r>>1);
        if(node[o].l==node[o].r){node[o]=Node(pos,pos,arr[pos]);return;}
        if(pos<=mid)chg(o<<1,pos);
        else chg(o<<1|1,pos);
        pushup(o);
    }
}segtree;
int main()
{
    ll n=read(), m=read(), i, j;
    rep(i,1,n)arr[i]=read();
    segtree.build(1,1,n);
    while(m--)
    {
        ll type=read();
        if(type==0)
        {
            ll l=read(), r=read();
            auto node = segtree.Sum(1,l,r);
            printf("%lld\n",node.best);
        }
        else
        {
            ll x=read(), v=read();
            arr[x]=v;
            segtree.chg(1,x);
        }
    }
    return 0;
}
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