HDU3450 Counting Sequences

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)

Total Submission(s): 1777 Accepted Submission(s): 596

Problem Description
For a set of sequences of integers｛a1，a2，a3，...an｝, we define a sequence｛ai1,ai2,ai3...aik｝in which 1<=i1= 2, and the neighboring 2 elements have the difference not larger than d, it will be defined as a Perfect Sub-sequence. Now given an integer sequence, calculate the number of its perfect sub-sequence.


Input
Multiple test cases The first line will contain 2 integers n, d（2<=n<=100000，1<=d=<=10000000） The second line n integers, representing the suquence


Output
The number of Perfect Sub-sequences mod 9901


Sample Input
4 2
1 3 7 5


Sample Output
4


Source
2010 ACM-ICPC Multi-University Training Contest（2）——Host by BUPT

题意：求给出串的子串中有相邻数字之差小于d的子串的个数。

分析：这题可以用树状数组和线段树做，两者的思路都差不多，首先都是对输入的数据离散化（d太大），然后都是以n为轴，建立树状数组（线段树），然后插入的时候要维护答案（DP），每次插入前先二分找出第一个大于等于a[i]-d所在的位置和最后一个小于等于a[i]+d所在的位置，然后只要查询这个区间就行了，插入是插入到a[i]所在的位置。

树状数组：

#include
#include
#include
using namespace std;
const int MAXN=100010;
const int MOD=9901;
int c[MAXN],hash[MAXN],a[MAXN];
int cnt;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    while(x=MOD)
            c[x]%=MOD;
        x+=lowbit(x);
    }
}
int getsum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=c[x];
        if(ans>=MOD)
            ans%=MOD;
        x-=lowbit(x);
    }
    return ans;
}
int search1(int x)
{
    int l=1,r=cnt,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(hash[mid]>1;
        if(hash[mid]<=x)
            l=mid+1;
        else
            r=mid-1;
    }
    return l-1;
}
int main()
{
    int n,d,i;
    while(scanf("%d%d",&n,&d)==2)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            hash[i]=a[i];
        }
        sort(hash+1,hash+1+n);
        cnt=1;
        for(i=2;i<=n;i++)
            if(hash[i]!=hash[cnt])
            hash[++cnt]=hash[i];
        int ans=0;
        memset(c,0,sizeof(c));
        for(i=1;i<=n;i++)
        {
            int y=search2(a[i]+d),x=search1(a[i]-d);
            int k=search1(a[i]);
            int sum=getsum(y)-getsum(x-1);
            sum=(sum+MOD)%MOD;
            ans=(ans+sum)%MOD;
            update(k,sum+1);
        }
        printf("%d\n",ans);
    }
    return 0;
}

线段树：

#include
#include
#include
#include
using namespace std;
const int MAXN=100010;
const int MOD=9901;
int cnt;
int a[MAXN],hash[MAXN];
struct seg
{
    int l,r;
    int num;
}tree[MAXN<<2];
void build(int l,int r,int k)
{
    tree[k].l=l;
    tree[k].r=r;
    tree[k].num=0;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(l,mid,k<<1);
    build(mid+1,r,k<<1|1);
}
void update(int n,int d,int k)
{
    if(tree[k].l==tree[k].r)
    {
        tree[k].num+=d;
        return;
    }
    int mid=(tree[k].l+tree[k].r)>>1;
    if(n<=mid)
        update(n,d,k<<1);
    else
        update(n,d,k<<1|1);
    tree[k].num=(tree[k<<1].num+tree[k<<1|1].num)%MOD;
}
int query(int l,int r,int k)
{
    if(tree[k].l>=l&&tree[k].r<=r)
        return tree[k].num;
    if(tree[k].l==tree[k].r)
        return tree[k].num;
    int mid=(tree[k].l+tree[k].r)>>1;
    if(r<=mid)
        return query(l,r,k<<1);
    else if(l>mid)
        return query(l,r,k<<1|1);
    else
        return query(l,mid,k<<1)+query(mid+1,r,k<<1|1);
}
int search1(int x)
{
    int l=1,r=cnt,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(hash[mid]>1;
        if(hash[mid]<=x)
            l=mid+1;
        else
            r=mid-1;
    }
    return l-1;
}
int main()
{
    int n,d,i;
    while(scanf("%d%d",&n,&d)==2)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            hash[i]=a[i];
        }
        sort(hash+1,hash+n+1);
        cnt=1;
        for(i=2;i<=n;i++)
            if(hash[cnt]!=hash[i])
            hash[++cnt]=hash[i];
        int ans=0;
        build(1,cnt,1);
        for(i=1;i<=n;i++)
        {
            int y=search2(a[i]+d),x=search1(a[i]-d);
            int k=search1(a[i]);
            int sum=query(x,y,1);
            sum=(sum+MOD)%MOD;
            ans=(ans+sum)%MOD;
            update(k,(sum+1)%MOD,1);
        }
        printf("%d\n",ans);
    }
    return 0;
}