鏈上問題是一個經典的貪心。於是考慮破環成鏈，將鏈倍長。求出每個線段右邊能作為後繼的最遠線段，然後倍增即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if
 
 (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 200010
int n,m,ans[N],f[N<<1][20];
struct data{int l,r,i;
}a[N<<1];
bool cmp(const data&a,const data&b)
{
    return a.l<b.l;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen(
 
"bzoj4444.in","r",stdin);
    freopen("bzoj4444.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++)
    {
        a[i].l=read(),a[i].r=read(),a[i].i=i;
        if (a[i].l>a[i].r) a[i].r+=m;
    }
    sort(a
 
+1,a+n+1,cmp);
    for (int i=n+1;i<=n*2;i++) a[i].l=a[i-n].l+m,a[i].r=min(m*2,a[i-n].r+m),a[i].i=a[i-n].i;
    int t=1;
    for (int i=1;i<=n*2;i++)
    {
        while (t<n*2&&a[t+1].l<=a[i].r) t++;
        f[i][0]=t;
    }
    for (int j=1;j<20;j++)
        for (int i=1;i<=n*2;i++)
        f[i][j]=f[f[i][j-1]][j-1];
    for (int i=1;i<=n;i++)
    {
        int x=i,cnt=1;
        for (int j=19;~j;j--) if (f[x][j]-i<n) cnt+=1<<j,x=f[x][j];
        ans[a[i].i]=cnt;
    }
    for (int i=1;i<=n;i++) printf("%d ",ans[i]);
    return 0;
}

BZOJ4444 SCOI2015國旗計劃（貪心+倍增）