UVALive - 7374 Racing Gems 二维非递减子序列

题目链接：

http://acm.hust.edu.cn/vjudge/problem/356795

Racing Gems

Time Limit: 3000MS
#### 问题描述
> You are playing a racing game. Your character starts at the
> x axis (y = 0) and proceeds up the race track, which has a
> boundary at the line x = 0 and another at x = w. You may
> start the race at any horizontal position you want, as long as
> it is within the track boundary. The finish line is at y = h,
> and the game ends when you reach that line. You proceed at
> a fixed vertical velocity v, but you can control your horizontal
> velocity to be any value between −v/r and v/r, and change it at any time.
> There are n gems at specific points on the race track. Your job is to collect as many gems as
> possible. How many gems can you collect?

输入

The input file contains several test cases, each of them as described below.

The first line of input contains four space-separated integers n, r, w, and h (1 ≤ n ≤ 105

, 1 ≤ r ≤ 10,

1 ≤ w, h ≤ 109

). Each of the following n lines contains two space-separated integers xi and yi

, denoting

the coordinate of the i-th gem (0 ≤ xi ≤ w, 0 < yi ≤ h). There will be at most one gem per location.

The input does not include a value for v.

输出

For each case, print, on a single line, the maximum number of gems that can be collected during the

race.

样例

sample input

5 1 10 10

8 8

5 1

4 6

4 7

7 9

5 1 100 100

27 75

79 77

40 93

62 41

52 45

10 3 30 30

14 9

2 20

3 23

15 19

13 5

17 24

6 16

21 5

14 10

3 6

sample output

3

3

4

题意

以辆赛车可以从x轴上任意点出发，他的水平速度允许他向每向上移动1个单位，就能向左或向右移动1/r个单位（也就是它的辐射范围是个等腰三角形）

现在赛车从x轴出发，问它在到达终点前能吃到的最多钻石。

题解

考虑每个钻石的向下辐射范围，并且将其投影到x轴上的两个点，（辐射范围与x轴的两个焦点），然后我们就把题目转化成了一个区间覆盖问题，我们用x<y表示x区间被y区间覆盖，即求二维最长的上升子序列：a1<=a2<=...<=an。

我们按每个钻石的左端点排序，然后跑右端点的最长不下降子序列就可以了。由于数据比较大，用二分处理成nlogn的复杂度。

代码

#include<map>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<string>
#define X first
#define Y second
#include<iostream>
#include<algorithm>
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define M (l+(r-l)/2)
#define bug(a) cout<<#a<<" = "<<a<<endl
using namespace std;

typedef __int64 LL;

const int maxn=1e5+10;

struct Node{
	LL a,b;
	bool operator <(const Node& tmp) const {
		return a>tmp.a||a==tmp.a&&b<tmp.b;
	}
}nds[maxn];

int n,w,h;
LL r;

LL ra[maxn];
int dp[maxn];

int main() {
	memset(dp,0,sizeof(dp));
	scanf("%d%I64d%d%d",&n,&r,&w,&h);
	for(int i=1;i<=n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		//映射到x轴的左端点
		nds[i].a=r*x-y;
		//映射到x轴的右端点
		nds[i].b=r*x+y;
	}
	//按左端点降序排，左端点相同时右端点升序排
	sort(nds+1,nds+n+1);
	//初始的ra数组为空
	int tot=1,ans=1;
	for(int i=1;i<=n;i++){
		//二分查找
		int pos=upper_bound(ra+1,ra+tot,nds[i].b)-ra;
		dp[i]=pos;
		ra[pos]=nds[i].b;
		if(tot==pos) tot++;
		ans=max(ans,dp[i]);
	}
	printf("%d\n",ans);
	return 0;
}