題目:
https://www.luogu.org/problem/show?pid=2587
Q
思路:
典型的田忌賽馬型貪心;
貪心政策:
盡可能讓強的赢;
對于a[],b[];
if a[]min > b[]min ,ans+=2;
else if a[]max>b[]max ,ans+=2;
else a[]min 和 b[]max 一組;
n*2是雙方得分的總分;
是以己方的最小得分=總分-對方的最大得分;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=;
int a[MAXN],b[MAXN],n;
int fi(int s1[],int s2[])
{
int l1=,l2=,r1=n,r2=n,ans=;
while(l1<=r1 && l2<=r2)
{
if(s1[l1]>s2[l2]) l1++,l2++,ans+=;
else if(s1[r1]>s2[r2]) r1--,r2--,ans+=;
else ans+=(s1[l1]==s2[r2]),l1++,r2--;
}
return ans;
}
void solve()
{
cin>>n;
for(int i=;i<=n;i++) scanf("%d",&a[i]);
for(int j=;j<=n;j++) scanf("%d",&b[j]);
sort(a+,a+n+),sort(b+,b+n+);
printf("%d %d",fi(a,b),(n<<)-fi(b,a));
}
int main()
{
solve();
return ;
}