2017百度之星初赛(B)-1006-小小粉丝度度熊 hdu 6119

去重合并，以某个点为起点，向后二分到最后一个满足未签到天数<=m的点。

二分时的查询有很多方法，可以用线段树，dp，前缀数组等等。比赛时没有想到前缀数组，就用来dp来完成查询，查询一次大约O(logn)。前缀数组的话查询O(1)更快一些。

同学说尺取法也可以。

dp[i][j]代表第i个区间到第i+(1<<j)个区间的间隙和

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))//0或-1
#define fs first
#define sc second
#define mp make_pair
#define L(x) (1<<x)
#define next Next
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=1e5+10;
const int maxm=1e3+10;
const int mod=1e9+7;
int n,m;
struct node
{
    int l,r;
}p[maxn];
bool cmp(const node &a,const node &b)
{
    return a.l<b.l;
}
int dp[maxn][20];
void init()
{
    for(int j=1;L(j)<=n;j++)
        for(int i=1;i+L(j)<=n;i++)
        dp[i][j]=dp[i][j-1]+dp[i+L(j-1)][j-1];
}
int query(int L,int R)
{
    if(L>=R) return 0;
    int t=0;
    while(1)
    {
        if(L+L(t+1)<=R)
            t++;
        else break;
    }
    return dp[L][t]+query(L+L(t),R);
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d %d",&p[i].l,&p[i].r);
        sort(p+1,p+1+n,cmp);
        int k=1;
        for(int i=2;i<=n;i++)
        {
            if(p[i].l<=p[k].r+1&&p[i].r>p[k].r)
                p[k].r=p[i].r;
            else if(p[i].l>p[k].r+1)
                p[++k]=p[i];
        }
        n=k;
        /*for(int i=1;i<=k;i++)
            printf("%d %d
",p[i].l,p[i].r);*/
        dp[n][0]=0;
        for(int i=1;i<=n-1;i++)
            dp[i][0]=p[i+1].l-p[i].r-1;
        init();
        int ans=m;
        for(int i=1;i<=n-1;i++)
        {
            int l=i+1,r=n;
            int temp=i;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(query(i,mid)<=m)
                {
                    temp=max(temp,mid);
                    l=mid+1;
                }
                else r=mid-1;
            }
            ans=max(ans,p[temp].r-p[i].l+1+m-query(i,temp));
        }
        ans=max(ans,p[n].r-p[n].l+1+m);
        printf("%d
",ans);
    }
    return 0;
}

sum[i]代表第i个区间之前的所有间隙和

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))//0或-1
#define fs first
#define sc second
#define mp make_pair
#define L(x) (1<<x)
#define next Next
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=1e5+10;
const int maxm=1e3+10;
const int mod=1e9+7;
int n,m;
struct node
{
    int l,r;
}p[maxn];
bool cmp(const node &a,const node &b)
{
    return a.l<b.l;
}
int sum[maxn];
int query(int L,int R)
{
    return sum[R]-sum[L];
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d %d",&p[i].l,&p[i].r);
        sort(p+1,p+1+n,cmp);
        int k=1;
        for(int i=2;i<=n;i++)
        {
            if(p[i].l<=p[k].r+1 && p[i].r>p[k].r)
                p[k].r=p[i].r;
            else if(p[i].l>p[k].r+1)
                p[++k]=p[i];
        }
        n=k;
        sum[1]=0;
        for(int i=2;i<=n;i++)
            sum[i]=sum[i-1]+p[i].l-p[i-1].r-1;
        int ans=m;
        for(int i=1;i<=n-1;i++)
        {
            int l=i+1,r=n;
            int temp=i;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(query(i,mid)<=m)
                {
                    temp=max(temp,mid);
                    l=mid+1;
                }
                else r=mid-1;
            }
            ans=max(ans,p[temp].r-p[i].l+1+m-query(i,temp));
        }
        ans=max(ans,p[n].r-p[n].l+1+m);
        printf("%d
",ans);
    }
    return 0;
}

 尺取法

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))//0或-1
#define fs first
#define sc second
#define mp make_pair
#define L(x) (1<<x)
#define next Next
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=1e5+10;
const int maxm=1e3+10;
const int mod=1e9+7;
int n,m;
struct node
{
    int l,r;
}p[maxn];
bool cmp(const node &a,const node &b)
{
    return a.l<b.l;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d %d",&p[i].l,&p[i].r);
        sort(p+1,p+1+n,cmp);
        int k=1;
        for(int i=2;i<=n;i++)
        {
            if(p[i].l<=p[k].r+1 && p[i].r>p[k].r)
                p[k].r=p[i].r;
            else if(p[i].l>p[k].r+1)
                p[++k]=p[i];
        }
        n=k;
        int l=1,r=1,cost=0;
        int ans=m;
        while(l<=n-1)
        {
            while(p[r+1].l-p[r].r-1 <= m-cost && r+1<=n)
            {
                cost+=p[r+1].l-p[r].r-1;
                r++;
            }
            ans=max(ans,p[r].r-p[l].l+1 + m-cost);
            if(l!=r)
            cost-=p[l+1].l-p[l].r-1;
            l++;
            r=max(r,l);
        }
        ans=max(ans,p[n].r-p[n].l+1 + m);
        printf("%d
",ans);
    }
    return 0;
}