CF1420D Rescue Nibel (greedy + thinking)

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Title:

have $n$ line segments$[l_i,r_i]$ , you need to select$k$ terms so that their intersection is not empty, find the number of$998244353$ modulo the result.

analyze:

After reading the question, we also know to find the intersection, then the intersection must be $max(l_i,l_j)<=min(r_i,r_j)$ so that the intersection is not empty, so we can first sort by the left node, then the right node. After sorting, we know that the left node will definitely increase. After we know that the left node is mono-increasing, we have actually finished comparing$max(l_i,l_j)$ , the left node in front of this node must be the largest. (Because we sort.)

Then we just need to see how many $r_j$(In the first (i-1)) is greater than or equal to l_i, which requires us to maintain a tree-like array, and put an $r_j$into a tree-like array.

/// 欲戴皇冠，必承其重。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef unsigned long long ull;

#define x first
#define y second
#define PI acos(-1)
#define inf 0x3f3f3f3f
#define lowbit(x) ((-x)&x)
#define debug(x) cout << #x << ": " << x << endl;

const int MOD = 998244353;
const int mod = 998244353;
const int N = 6e5 + 10;
const int dx[] = {
    
    0, 1, -1, 0, 0, 0, 0};
const int dy[] = {
    
    0, 0, 0, 1, -1, 0, 0};
const int dz[] = {
    
    0, 0, 0, 0, 0, 1, -1};
int day[] = {
    
    0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

ll n, m;
ll fac[N],infac[N];
struct node 
{
    
    
    ll l,r;
}q[N];
bool cmp(node a,node b){
    
    
    if(a.l!=b.l)
    return a.l<b.l;
    return a.r<b.r;
}
ll qpow(ll a, ll b){
    
    
    ll ans=1;
    while(b){
    
    
        if(b&1) ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
ll a[N],b[N];
ll num;
void add(ll x){
    
    
    while(x<=num){
    
    
        b[x]++;
        x+=lowbit(x);        
    }    
}
ll ask(ll x){
    
    
    ll ans=0;
    while(x){
    
    
        ans+=b[x];
        x-=lowbit(x);
    }
    return ans;
}
ll C(ll x,ll y){
    
    
    return fac[x]*infac[y]%mod*infac[x-y]%mod;
}
void solve(){
    
        
    scanf("%lld%lld",&n,&m);
    ll ans=0;
    fac[0]=infac[0]=1;
    for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
    for(int i=1;i<=n;i++) infac[i]=qpow(fac[i],mod-2);
    for(int i =1;i<=n;i++){
    
    
        scanf("%lld%lld",&q[i].l,&q[i].r);
        a[++num]=q[i].l;
        a[++num]=q[i].r;
    }
    sort(q+1,q+1+n,cmp);
    sort(a+1,a+1+num);
    num=unique(a+1,a+1+num)-a-1;
    for(int i=1;i<=n;i++){
    
    
        q[i].l=lower_bound(a+1,a+1+num,q[i].l)-a;
        q[i].r=lower_bound(a+1,a+1+num,q[i].r)-a;
        ll cnt = ask(num)-ask(q[i].l-1);//ll cnt = (i-1)-ask(q[i].l-1);
        add(q[i].r);
        if(cnt>=m-1){
    
    
            ans=(ans+C(cnt,m-1))%mod;
        }        
    }
    printf("%lld\n",ans);
}
int main()
{
    
    
    ll t = 1;
    //scanf("%lld", &t);
    while(t--)
    {
    
    
        solve();
    }
    return 0;
}