【刷题】BZOJ 4391 [Usaco2015 dec]High Card Low Card

Description

Bessie the cow is a huge fan of card games, which is quite surprising, given her lack of opposable thumbs. Unfortunately, none of the other cows in the herd are good opponents. They are so bad, in fact, that they always play in a completely predictable fashion! Nonetheless, it can still be a challenge for Bessie to figure out how to win.

Bessie and her friend Elsie are currently playing a simple card game where they take a deck of 2N cards, conveniently numbered 1…2N, and divide them into N cards for Bessie and N cards for Elsie. The two then play NN rounds, where in each round Bessie and Elsie both play a single card. Initially, the player who plays the highest card earns a point. However, at one point during the game, Bessie can decide to switch the rules so that for the rest of the game, the player who plays the lowest card wins a point. Bessie can choose not to use this option, leaving the entire game in "high card wins" mode, or she can even invoke the option right away, making the entire game follow the "low card wins" rule.

Given that Bessie can predict the order in which Elsie will play her cards, please determine the maximum number of points Bessie can win.

奶牛Bessie和Elsie在玩一种卡牌游戏。一共有2N张卡牌，点数分别为1到2N，每头牛都会分到N张卡牌。

游戏一共分为N轮，因为Bessie太聪明了，她甚至可以预测出每回合Elsie会出什么牌。

每轮游戏里，两头牛分别出一张牌，点数大者获胜。

同时，Bessie有一次机会选择了某个时间点，从那个时候开始，每回合点数少者获胜。

Bessie现在想知道，自己最多能获胜多少轮？

Input

The first line of input contains the value of N (2≤N≤50,000).

The next N lines contain the cards that Elsie will play in each of the successive rounds of the game. Note that it is easy to determine Bessie's cards from this information.

Output

Output a single line giving the maximum number of points Bessie can score.

Sample Input

4

1

8

4

3

Sample Output

3

HINT

Here, Bessie must have cards 2, 5, and 6, and 7 in her hand, and she can use these to win at most 3 points. For example, she can defeat the 1 card and then switch the rules to "low card wins", after which she can win two more rounds.

Solution

怎么说，贪心的题都不知道怎么去讲

首先，对于单个数，我们要赢它，那么肯定是在自己的数的集合中选一个大于它的最小的数，或者是小于它的最大的数（取决在哪一段）

这是单个数的贪心

然后求出 \(f\) 数组，\(f_i\) 代表从前往后到第 \(i\) 个数，赢的方式必须是大于的情况下，最多能赢多少把

同时求出 \(g\) 数组，\(g_i\) 代表从后往前到第 \(i\) 个数，赢的方式必须是小于的情况下，最多能赢多少把

那么最后的答案就是 \(\max_{i=0}^n\{f_i+g_{i+1}\}\)

为什么是这样，这样有的数不是用了两次吗

确实有的数用了两次，但考虑这样的事实：

• 如果一个数用了两次，那么肯定有一个数未被使用。

• 用了两次的数一定是 \(f\) 中用一次，\(g\) 中用一次，同一段中不可能用两次

设用了两次的数为 \(a\) ，未用的数为 \(b\) 。如果 \(a<b\) ，那么 \(b\) 是可以代替 \(a\) 在比更大的那一段取得胜利的；相反同理。

而我们已经最大化了 \(f\) 数组和 \(g\) 数组，所以这一定是正确的

求两个数组，就用个set维护就好了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=50000+10;
int n,vis[MAXN<<1],a[MAXN],f[MAXN],g[MAXN],cnt,ans;
std::set<int> S;
std::set<int>::iterator it;
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
int main()
{
	read(n);
	for(register int i=1;i<=n;++i)read(a[i]),vis[a[i]]=1;
	for(register int i=1;i<=(n<<1);++i)
		if(!vis[i])S.insert(i);
	for(register int i=1;i<=n;++i)
		if((it=S.lower_bound(a[i]))!=S.end())f[i]=++cnt,S.erase(it);
		else f[i]=f[i-1];
	S.clear();cnt=0;
	for(register int i=1;i<=(n<<1);++i)
		if(!vis[i])S.insert(i);
	for(register int i=n;i>=1;--i)
		if(((it=S.lower_bound(a[i]))--)!=S.begin())g[i]=++cnt,S.erase(it);
		else g[i]=g[i+1];
	for(register int i=0;i<=n;++i)chkmax(ans,f[i]+g[i+1]);
	write(ans,'\n');
	return 0;
}