4396: [Usaco2015 dec]High Card Wins
Time Limit: 10 Sec
Memory Limit: 128 MB
Submit: 31
Solved: 23
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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
2
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
贪心。将两个人排序，用a和b两个数组表示。每次从a数组中拿出一个最小的，再从b数组中拿出一个最小的大于这个数的数，直到b数组中所有数都小于a数组中的数。（额...大家理解就行了）
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define pa pair<int,int>
#define maxn 100005
#define inf 1000000000
using namespace std;
int n,x,cnta,cntb,j=0,ans=0;
int a[maxn],b[maxn];
bool f[maxn];
inline int read()
{
int x=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int main()
{
n=read();
F(i,1,n){x=read();f[x]=true;}
F(i,1,2*n)
{
if (f[i]) a[++cnta]=i;
else b[++cntb]=i;
}
F(i,1,n)
{
j++;
while (j<=n&&b[j]<a[i]) j++;
if (j>n) break;
ans++;
}
printf("%d\n",ans);
}
