4396: [Usaco2015 dec]High Card Wins

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 275 Solved: 175
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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 N rounds, where in each round Bessie and Elsie both play a single card, and the player with the highest card earns a point.
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。

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 (1≤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

3
1
6
4

Sample Output

HINT

Here, Bessie must have cards 2, 3, and 5 in her hand, and she can use these to win at most 2 points by saving the 5 until the end to beat Elsie's 4.

Problem credits: Austin Bannister and Brian Dean

Source

Silver鸣谢Claris提供译文

set似乎可以强搞，利用set自排序的特性，再按照田忌赛马原则（？）大雾

 1 #include
 2 #include
 3 #include<set>
 4 #include
 5 #include
 6 using namespace std;
 7 
 8 int n,node=0;
 9 int a[50005];
10 set<int> S;
11 set<int>::iterator iter;
12 
13 int main()
14 {
15     scanf("%d",&n);
16     for(int i=1;i<=2*n;i++) S.insert(i);
17     for(int i=0;i)
18     {
19         scanf("%d",&a[i]);
20         S.erase(a[i]);
21     }
22     sort(a,a+n);
23     for(iter=S.begin();iter!=S.end();iter++)
24         if(*iter>a[node]) node++;
25     printf("%d",node);
26     return 0;
27 }