The coordinates are discretized first, because the intervals are not contained, and can be understood as the minimum number of fighters required for each start output.
Will ring times long, broken loop into a chain, set $f[i]$ to indicate that the interval left endpoint does not exceed the $i$ when the right end of the maximum value, can be deduced by $o (n) $ recursion.
So if $f[i]$ is seen as an ancestor of $i$, it actually forms a tree rooted in $2n$.
First of all, the violence calculated the answer of number 1th, $t$, set $l=t-1$.
Then DFS this tree, using a stack to store each point to a point on the root path in order of depth.
For a point, you only need to start with $l$ to enumerate the answers violently, and then $o (1) $ test.
Because the answer of each point is not more than 1, the time complexity of the whole algorithm is $o (n) $ after the discretization is removed.
#include <cstdio> #include <algorithm> #define N 400010int n,m,i,x,y,l,a[n/2][2],b[n],st[n/2],f[n*2],g[n* 2],nxt[n*2],q[n*2],t,ans[n];inline void Read (int&a) {char c;while (!) ( ((C=getchar ()) >= ' 0 ') && (c<= ' 9 ')); a=c-' 0 '; while (((C=getchar ()) >= ' 0 ') && (c<= ' 9 ')) (a*= Ten) +=c-' 0 ';} inline int lower (int x) {int l=1,r=m,mid,t; while (l<=r) if (b[mid= (l+r) >>1]<=x) l= (t=mid) +1;else r=mid-1; return t;} inline void up (Int&x,int y) {if (x<y) x=y;} void Dfs (int x) {q[++t]=x; if (x<=m) for (int i=l;; i++) if (q[t-i]>=x+m) {ans[x]=i;break;} for (int i=g[x];i;i=nxt[i]) DFS (i); t--;} int main () {read (n), read (m); for (m=0,i=1;i<=n;i++) read (a[i][0]), read (a[i][1]), b[++m]=a[i][0],b[++m]=a[i][1]; For (Std::sort (b+1,b+m+1), i=1;i<=n;i++) {st[i]=x=lower (a[i][0]), Y=lower (a[i][1]); if (x<y) up (F[x],y), Up (f[x+m],y+m); else up (F[1],y), Up (F[x],y+m), Up (f[x+m],m+m); } for (i=1;i<=m+m;i++) up (f[i],f[i-1]); for (i=1;i<m+m;i++) nxt[i]=g[f[i]],g[f[i]]=i; For(L=-1,i=1;i<=m;i=f[i]) l++; DFS (M+M); for (i=1;i<=n;i++) printf ("%d", Ans[st[i]]); return 0;}
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