HDOJ 题目2836 Traversal（线段树，离散化，DP）

Traversal

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 812    Accepted Submission(s): 301

Problem Description

I arrive at a big lake and I have to cross it. Luckily, I’m a very good jumper, but the lake is too big to be crossed in one jump. On the shore, I find N boxes of different heights, set in a certain order. If I throw a box into the lake, it will float and it will have the same height as on the shore. This is good, because I intend to throw some boxes into the lake and get from one shore to the other by jumping from box to box. The only things to consider are:
The lake is big, so I must throw at least 2 boxes, which means that in order to cross the lake I have to make at least 3 jumps.
Not all the boxes have to be thrown; some of them may be ignored.
The boxes can be thrown into the lake only in the order they are found on the shore and I have to jump on them in this order.
The height difference between two consecutive boxes I use must be at most H meters, because I can jump a lot in length, but I have some problems with jumping in height.The height of a box doesn’t change when I jump on it.
I’m always able to jump from the shore to a box and from a box to the shore, no matter what the height of the box is.

Facing so many possibilities that respect the above conditions, I begin counting the number of possibilities that I have, instead of actually crossing the lake. I quickly find the answer and I wonder whether you can also find it as fast as I did.

Task

Write a program that determines the number of possibilities to cross the lake in the above conditions. Since the number can be quite big, you only have to output the remainder of this number, when divided by 9901.

 

Input

There are multiple test cases. Each test case contains two integers N and H, separated by a space, representing the number of boxes and the maximum height difference between two consecutive boxes thrown into the lake. The following N lines contain the heights of the boxes, in the order the boxes are set on the shore. The (i+1)th line contains the height of the ith box.

 

Output

For each test case you should output a single line, containing the number of possibilities modulo 9901.

Constraints

1 < N < 100 001
0 < H < 100 000 001
The height of any box is a strictly positive integer and does not exceed 100 000 000

 

Sample Input

4 21 3 7 5

 

Sample Output

4
Hint
ExplanationThere are 4 possibilities:1 31 3 53 57 5
 

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

Recommend

gaojie   |   We have carefully selected several similar problems for you:  2835 2843 2841 2839 2837 

 

题目大意：给个n，m,就是从一个序列中找一个序列，使相邻的两个数的差不大于m

ac代码

153641052015-11-03 16:13:48Accepted2836249MS4948K2761 BG++XY_

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>//#define mod 9901#define LL long longusing namespace std;LL a[100010],b[100010];LL node[100010<<2],n;LL dp[100010];void pushup(int tr){    node[tr]=(node[tr<<1]+node[tr<<1|1])%9901;}void build(int l,int r,int tr){    node[tr]=0;    if(l==r)    {        return;    }    int mid=(l+r)>>1;    build(l,mid,tr<<1);    build(mid+1,r,tr<<1|1);    //pushup(tr);}void update(int pos,LL val,int l,int r,int tr){    if(l==r)    {        node[tr]=(node[tr]+val);        return;    }    int mid=(l+r)>>1;    if(pos<=mid)        update(pos,val,l,mid,tr<<1);    else        update(pos,val,mid+1,r,tr<<1|1);    pushup(tr);}LL query(int L,int R,int l,int r,int tr){    if(L<=l&&r<=R)    {        return node[tr];    }    int mid=(l+r)>>1;    if(R<=mid)        return query(L,R,l,mid,tr<<1);    else        if(L>mid)            return query(L,R,mid+1,r,tr<<1|1);        else        {            int a,b;            a=query(L,mid,l,mid,tr<<1);            b=query(mid+1,R,mid+1,r,tr<<1|1);            return (a+b)%9901;        }}int bseach(LL val){    int l=1;    int r=n;    while(l<=r)    {        int mid=(l+r)>>1;        if(b[mid]==val)            return mid;        if(b[mid]<val)            l=mid+1;        else            r=mid-1;    }    return l;}int bseach_r(LL val){    int l=1;    int r=n;    int ans;    while(l<=r)    {        int mid=(l+r)>>1;        if(b[mid]<=val)        {            ans=mid;            l=mid+1;        }        else            r=mid-1;    }    return ans;}int bseach_l(LL val){    int l=1,r=n;    int ans;    while(l<=r)    {        int mid=(l+r)>>1;        if(b[mid]>=val)        {            ans=mid;            r=mid-1;        }        else            l=mid+1;    }    return ans;}int main(){    int N,m;    while(scanf("%d%d",&N,&m)!=EOF)    {        int i;        memset(dp,0,sizeof(dp));        for(i=1;i<=N;i++)        {            scanf("%lld",&a[i]);            b[i]=a[i];           // node[i]=0;        }        sort(b+1,b+1+N);        n=unique(b+1,b+1+N)-b-1;        build(1,n,1);        dp[1]=0;        update(bseach(a[1]),dp[1]+1,1,n,1);        LL ans=0;        for(i=2;i<=N;i++)        {            int l=bseach_l(a[i]-m);            int r=bseach_r(a[i]+m);           // printf("%d %d\n",l,r);            dp[i]=query(l,r,1,n,1);            int x=bseach(a[i]);            ans=(ans+dp[i])%9901;            update(x,dp[i]+1,1,n,1);        }        printf("%lld\n",ans);    }}