洛谷P3253 [JLOI2013]删除物品

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题解

长时间没做算法题，连这个水题都要做好久qwq

这个题就是模拟这样的一个过程：

每次把最大的那个数上方的数字弹到另一个栈里面，然后把这个最大数删去

我就头对头把这两个栈拼接起来，然后维护栈顶的位置

拿数据结构算一下每次弹出了多少次就行了

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 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 BIT
{
    ll bit[maxn], n;
    void init(ll N){n=N;for(ll i=1;i<=n;i++)bit[i]=0;}
    ll lowbit(ll x){return x&(-x);}
    void add(ll pos, ll v)
    {
        for(;pos<=n;pos+=lowbit(pos))bit[pos]+=v;
    }
    ll sum(ll pos)
    {
        ll ans(0);
        for(;pos;pos-=lowbit(pos))ans+=bit[pos];
        return ans;
    }
}bit;
struct Lisan
{
    ll tmp[maxn], tot;
    void clear(){tot=0;}
    void insert(ll x){tmp[++tot]=x;}
    void run()
    {
        sort(tmp+1,tmp+tot+1);
        tot=unique(tmp+1,tmp+tot+1)-tmp-1;
    }
    void lisan(ll *a, ll len)
    {
        for(ll i=1;i<=len;i++)a[i]=lower_bound(tmp+1,tmp+tot+1,a[i])-tmp;
    }
    ll lisan(ll x)
    {
        return lower_bound(tmp+1,tmp+tot+1,x)-tmp;
    }
}ls;
ll n1, n2, n, a[maxn], ans, pos[maxn], now;
int main()
{
    ll i, add;
    n1=read(), n2=read();
    n = n1+n2;
    drep(i,n1,1)a[i]=read(), ls.insert(a[i]);
    rep(i,n1+1,n)a[i]=read(), ls.insert(a[i]);
    ls.run();
    ls.lisan(a,n);
    now=n1;
    rep(i,1,n)pos[a[i]]=i;
    bit.init(n);
    rep(i,1,n)bit.add(i,+1);
    drep(i,n,1)
    {
        if(pos[i]>now)
        {
            ans += bit.sum(pos[i]-1) - bit.sum(now);
            now = pos[i];
        }
        else
        {
            ans += bit.sum(now) - bit.sum(pos[i]);
            now = pos[i];
        }
        bit.add(pos[i],-1);
    }
    printf("%lld",ans);
    return 0;
}