建设企业网站支票打印软件/天津seo优化排名

https://atcoder.jp/contests/arc165/tasks/arc165_f

首先可以建图，然后变成求字典序最小的的拓扑排序

然后发现这样复杂度会炸，观察连边的条件是什么：

• $l_i<l_j$
• $r_i<r_j$

这是个二维偏序问题，我们考虑用分治来解决

我们按 $l$ 排序，本区间内再按 $r$ 排序：

复杂度 $O(n{\log }^{2}n)$

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else#define debug(...) void(0)
#endif
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//srand(time(0));
#define N 5000010
//#define M
//#define mo
struct node {int l, r, id, op; 
}a[N];
int n, m, i, j, k, T;
struct cmp {bool operator () (int x, int y) const {if(x<=n && y>n) return 1; if(y<=n && x>n) return 0; if(x<=n && y<=n) return x>y; if(x>n && y>n) return x>y; }
};
priority_queue<int, vector<int>, cmp >q; 
int c[N], b[N], tot; 
vector<int>G[N]; void cun(int x, int y) {
//	debug("%d -> %d\n", x, y); G[x].pb(y); ++c[y]; 
}void solve(int l, int r) {int i; if(l==r) return ; int mid=(l+r)>>1; solve(l, mid); solve(mid+1, r); for(i=l; i<=mid; ++i) a[i].op=0; for(i=mid+1; i<=r; ++i) a[i].op=1; sort(a+l, a+r+1, [&] (node x, node y) { return x.r<y.r; }); for(i=l; i<=r; ++i) b[i]=++tot; for(i=l; i<=r-1; ++i) cun(b[i], b[i]+1); for(i=l; i<=r; ++i) if(a[i].op==0) cun(a[i].id, b[i]); else cun(b[i], a[i].id); 
}signed main()
{#ifdef LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif
//	T=read();
//	while(T--) {
//
//	}n=read(); tot=n; for(i=1; i<=n; ++i) a[i].id=i; for(i=1; i<=2*n; ++i) {k=read(); if(!a[k].l) a[k].l=i; else a[k].r=i; }sort(a+1, a+n+1, [] (node x, node y) { return x.l<y.l; }); solve(1, n); for(i=1; i<=tot; ++i) if(!c[i]) q.push(i); while(!q.empty()) {auto u=q.top(); q.pop(); 
//		debug("# %d\n",  u); if(u<=n) printf("%d %d ", u, u); for(auto v : G[u]) if(--c[v]==0) q.push(v); }return 0;
}