HDU2217Visit

Visit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 644 Accepted Submission(s): 250

Problem Description

Wangye is interested in traveling. One day, he want to make a visit to some
different places in a line. There are N(1 <= N <= 2000) places located at points x1, x2, …, xN (-100,000 ≤ xi ≤ 100,000). Wangye starts at HDU (x = 0), and he travels 1 distance unit in 1 minute . He want to know how many places he could visit at most, if he has T (1 <= T <= 200000 ) minutes.

Input

The input contains several test cases .Each test case starts with two number N and T which indicate the number of places and the time respectively. Then N lines follows, each line has a number xi indicate the position of i-th place.

Output

For each test case, you should print the number of places,Wangye could visit at most, in one line.

Sample Input

5 16
-3
-7
1
10
8

Sample Output

4
Hint

In the sample, Wangye could visit (-3) first, then goes back to (0), which costs him 6 minutes.
Then he visits (1), (8), (10). So he could visit 4 places at most in 16 minutes. :

Author

zjt
题意：给你n个点和t时间和每个点的坐标，问在t时间内最多能经过多少个点。
题解：
将点分成正负两个集合。枚举每一个点作为端点的情况，记录最大值即可。这里用了一个前缀和，记录距离<=i有多少个点。
注意：多点在一个地方的情况和初始点（0）是否有点的情况
代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <math.h>#include <algorithm>#include <string.h>using namespace std;const int N=1e6+10;int l[N],r[N];int a[N],b[N];int n,t;int main(){    while(~scanf("%d%d",&n,&t))    {        memset(l,0,sizeof(l));         memset(r,0,sizeof(r));          memset(a,0,sizeof(a));          memset(b,0,sizeof(b));        int tmp,tmpp,x;        tmp=tmpp=1;        int sum=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&x);            if(x>0)            {                l[tmp++]=x;                a[x]++;            }            else if(x<0)            {                r[tmpp++]=-x;                b[-x]++;            }            else            {                sum++;            }        }        sort(l+1,l+tmp);        sort(r+1,r+tmpp);        for(int i=1;i<N;i++)        {            a[i]+=a[i-1];            b[i]+=b[i-1];        }        int ans=0;        for(int i=0;i<tmpp;i++)        {            if(t-2*r[i]>=0)            ans=max(ans,a[t-2*r[i]]+b[r[i]]);        }        for(int i=0;i<tmp;i++)        {            if(t-2*l[i]>=0)            ans=max(ans,b[t-2*l[i]]+a[l[i]]);        }            ans+=sum;        printf("%d\n",ans);    }}

#include <iostream>#include<stdio.h>#include<algorithm>using namespace std;int n,T,a[2005];int main(){    int s,q,p,ma,x,y;    while(scanf("%d %d",&n,&T)!=EOF)    {        a[0]=0;        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        sort(a,a+n+1);        s=lower_bound(a,a+n+1,0)-a;        ma=0;        for(q=s;q>=0;q--)        {            for(p=s;p<=n;p++)            {                x=-a[q],y=a[p];                if(2*x+y<=T&&p-q>ma) ma=p-q;                if(2*y+x<=T&&p-q>ma) ma=p-q;            }        }        cout<<ma<<endl;    }    return 0;}