cf1327F. AND Segments

Link

Link

Solution

dp

拆位做

现在相当于有些区间必须全是 $1$

有些区间至少有一个 $0$，我把这个叫做零区间

$f_i$表示位于 $[1,i)$的零区间都满足了条件，且第 $i$个位置上是 $0$的方案数

Code

#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 500010
#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;
}
#define mod 998244353ll
ll f[maxn], n, k, m, a[maxn], x[maxn], l[maxn], r[maxn], id[maxn], s[maxn], sf[maxn];
vector<pll> seg;
ll dp()
{
    auto it = seg.begin();
    ll i, leftmost=0;
    sf[0]=f[0]=1;
    rep(i,1,n+1)
    {
        if(s[i])f[i]=0;
        else
        {
            while(it!=seg.end() and it->second<i)
            {
                leftmost = max( leftmost, it->first );
                it++;
            }
            f[i] = (sf[i-1] - sf[leftmost-1]) %mod;
        }
        sf[i]=sf[i-1]+f[i];
    }
    return f[n+1];
}
int main()
{
    ll i, j, ans=1;
    n=read(), k=read(), m=read();
    rep(i,1,m)l[i]=read(), r[i]=read(), x[i]=read(), id[i]=i;
    sort(id+1,id+m+1,[](ll x, ll y){return r[x]<r[y];});
    rep(j,0,k-1)
    {
        seg.clear();
        rep(i,1,n+1)s[i]=0, f[i]=0;
        rep(i,1,m)
        {
            ll q=id[i];
            if(x[q]&(1<<j))s[l[q]]++, s[r[q]+1]--;
            else seg.emb(pll(l[q],r[q]));
        }
        rep(i,1,n+1)s[i]+=s[i-1];
        ans = ans * dp() %mod;
    }
    printf("%lld",(ans+mod)%mod);
    return 0;
}