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 ;

}