The main idea is to put right in front of discrete points ++, right after discrete point 1.
1. CF 610D Vika and Segments
Effect: a given segment $ $ n-coordinate axis parallel to the strip, the number of points required together represent
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 4e5+10; int n, cas, cnt, tot; struct _ {int l,r,h,v;} e[N]; int b[N], sum[N<<2]; int tag[N<<2]; void pu(int o, int l, int r) { if (tag[o]) sum[o] = b[r+1]-b[l]; else sum[o] = sum[lc]+sum[rc]; } void update(int o, int l, int r, int ql, int qr, int v) { if (ql<=l&&r<=qr) return tag[o]+=v,pu(o,l,r); if (mid>=ql) update(ls,ql,qr,v); if (mid<qr) update(rs,ql,qr,v); pu(o,l,r); } int main() { scanf("%d", &n); REP(i,1,n) { int x1, x2, y1, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); if (x1>x2) swap(x1,x2); if (y1>y2) swap(y1,y2); --x1,--y1; e[++cnt]={y1,y2,x1,1}; e[++cnt]={y1,y2,x2,-1}; b[++tot]=y1,b[++tot]=y2; } sort(b+1,b+1+tot),tot=unique(b+1,b+1+tot)-b-1; sort(e+1,e+1+cnt,[](_ a,_ b){return a.h<b.h;}); ll ans = 0; REP(i,1,cnt-1) { int l=lower_bound(b+1,b+1+tot,e[i].l)-b; int r=lower_bound(b+1,b+1+tot,e[i].r)-b-1; update(1,1,tot,l,r,e[i].v); ans += (ll)sum[1]*(e[i+1].h-e[i].h); } printf("%lld\n", ans); }
2. CF 817F MEX Queries
Effect: Flip the interval, the interval assignment, seeking $ 0 $ leftmost position
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 2e5+10; int n, cnt, T, tot; ll b[N]; struct _ { int tag,v[2]; void upd(int x,int y) { if (y==1) v[0]=cnt,v[1]=x; else v[0]=x,v[1]=cnt; } void flip() {swap(v[0],v[1]);tag^=1;} } tr[N<<2]; void pd(int o, int l, int r) { if (tr[o].tag) { tr[lc].flip(); tr[rc].flip(); tr[o].tag = 0; } if (tr[o].v[0]==cnt) { tr[lc].upd(l,1); tr[rc].upd(mid+1,1); } if (tr[o].v[1]==cnt) { tr[lc].upd(l,0); tr[rc].upd(mid+1,0); } } void add(int o, int l, int r, int ql, int qr, int v) { if (ql<=l&&r<=qr) return tr[o].upd(l,v); pd(o,l,r); if (mid>=ql) add(ls,ql,qr,v); if (mid<qr) add(rs,ql,qr,v); tr[o].v[0]=min(tr[lc].v[0],tr[rc].v[0]); tr[o].v[1]=min(tr[lc].v[1],tr[rc].v[1]); } void flip(int o, int l, int r, int ql, int qr) { if (ql<=l&&r<=qr) return tr[o].flip(); pd(o,l,r); if (mid>=ql) flip(ls,ql,qr); if (mid<qr) flip(rs,ql,qr); tr[o].v[0]=min(tr[lc].v[0],tr[rc].v[0]); tr[o].v[1]=min(tr[lc].v[1],tr[rc].v[1]); } void build(int o, int l, int r) { tr[o].upd(l,0); if (l!=r) build(ls),build(rs); } struct {char op;ll l,r;} q[N]; int main() { scanf("%d", &n); REP(i,1,n) { int op; ll l,r; scanf("%d%lld%lld",&op,&l,&r); q[i].op=op,q[i].l=l,q[i].r=r; b[++cnt]=q[i].l; b[++cnt]=++q[i].r; } b[++cnt]=1,b[++cnt]=2e18; sort(b+1,b+1+cnt),cnt=unique(b+1,b+1+cnt)-b-1; build(1,1,cnt-1); REP(i,1,n) { q[i].l=lower_bound(b+1,b+1+cnt,q[i].l)-b; q[i].r=lower_bound(b+1,b+1+cnt,q[i].r)-b-1; if (q[i].op<=2) add(1,1,cnt-1,q[i].l,q[i].r,q[i].op==1); else flip(1,1,cnt-1,q[i].l,q[i].r); printf("%lld\n",b[tr[1].v[0]]); } }