BZOJ4444 SCOI2015国旗计划（贪心+倍增）

　　链上问题是一个经典的贪心。于是考虑破环成链，将链倍长。求出每个线段右边能作为后继的最远线段，然后倍增即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
int n,m,ans[N],f[N<<][];
struct data{int l,r,i;
}a[N<<];
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=;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+,a+n+,cmp);
    for (int i=n+;i<=n*;i++) a[i].l=a[i-n].l+m,a[i].r=min(m*,a[i-n].r+m),a[i].i=a[i-n].i;
    int t=;
    for (int i=;i<=n*;i++)
    {
        while (t<n*&&a[t+].l<=a[i].r) t++;
        f[i][]=t;
    }
    for (int j=;j<;j++)
        for (int i=;i<=n*;i++)
        f[i][j]=f[f[i][j-]][j-];
    for (int i=;i<=n;i++)
    {
        int x=i,cnt=;
        for (int j=;~j;j--) if (f[x][j]-i<n) cnt+=<<j,x=f[x][j];
        ans[a[i].i]=cnt;
    }
    for (int i=;i<=n;i++) printf("%d ",ans[i]);
    return ;
}