真题补题笔记

P11830 [省选联考 2025] 幸运数字

后面忘了，随机分讨，讨论出每一段前面，当前，后面的可能数量区间，还需要对离散化后每个点和两个点之间的段进行分讨，应该是写烦了

#include<bits/stdc++.h>
#define fi first
#define se second
#define N 400005
#define int long long
using namespace std;
struct Ty{
    int l1,r1,l2,r2;
    bool operator <(const Ty &a)const{return l2<a.l2;}
}x[N];
priority_queue<pair<int,int> >q;
int z[N],n,m=0,prel[N],prer[N],sufl[N],sufr[N],xl[N],xr[N],pxl[N],pxr[N];
signed main(){
    int c,T;
    scanf("%lld%lld",&c,&T);
    while(T--){
        m=0;
        scanf("%lld",&n);
        for(int i=1;i<=n;i++)scanf("%lld%lld%lld%lld",&x[i].l1,&x[i].r1,&x[i].l2,&x[i].r2);
        for(int i=1;i<=n;i++){
            z[++m]=x[i].l2;
            z[++m]=x[i].r2;
        }
        sort(z+1,z+m+1);
        m=unique(z+1,z+m+1)-z-1;
        for(int i=1;i<=n;i++){
            x[i].l2=lower_bound(z+1,z+m+1,x[i].l2)-z;
            x[i].r2=lower_bound(z+1,z+m+1,x[i].r2)-z;
        }
        sort(x+1,x+n+1);
        int now=1,pl=0,pr=0,l=0,r=0;
        while(!q.empty())q.pop();
        for(int i=1;i<=m+1;i++){
            pxl[i]=l;
            pxr[i]=r;
            while(now<=n&&x[now].l2<=i){
                l+=x[now].l1;
                r+=x[now].r1;
                q.push({-x[now].r2,now});
                now++;
            }
            xl[i]=l;
            xr[i]=r;
            while(!q.empty()&&-q.top().fi<=i){
                int u=q.top().se;
                pl+=x[u].l1;
                pr+=x[u].r1;
                l-=x[u].l1;
                r-=x[u].r1;
                q.pop();
            }
            prel[i]=pl;
            prer[i]=pr;
        }
        now=n;
        pl=pr=l=r=0;
        while(!q.empty())q.pop();
        for(int i=m+1;i;i--){
            while(now&&x[now].r2>=i){
                l+=x[now].l1;
                r+=x[now].r1;
                q.push({x[now].l2,now});
                now--;
            }
            while(!q.empty()&&q.top().fi>=i){
                int u=q.top().se;
                pl+=x[u].l1;
                pr+=x[u].r1;
                l-=x[u].l1;
                r-=x[u].r1;
                q.pop();
            }
            sufl[i]=pl;
            sufr[i]=pr;
        }
        int ans=0;
        for(int i=1;i<=m;i++){
            if(prer[i-1]>=sufl[i]&&prel[i-1]<=sufr[i]&&pxr[i]>0)ans+=z[i]-z[i-1]-1;
            else if(pxr[i]>0){
                if(prer[i-1]<sufl[i]){
                    if(prer[i-1]+pxr[i]>=sufl[i])ans+=z[i]-z[i-1]-1;
                }
                else{
                    if(sufr[i]+pxr[i]>prel[i-1])ans+=z[i]-z[i-1]-1;
                }
            }
            if(prer[i-1]>=sufl[i+1]&&prel[i-1]<=sufr[i+1]&&xr[i]>0)ans++;
            else if(xr[i]>0){
                if(prer[i-1]<sufl[i+1])ans+=(prer[i-1]+xr[i]>=sufl[i+1]);
                else ans+=(sufr[i+1]+xr[i]>prel[i-1]);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}