假設(shè)現(xiàn)在有 k 個(gè) “2020”，也就是需要選擇 2 * k 個(gè) “20” 進(jìn)行匹配，一種較優(yōu)的貪心方案就是，從左邊開(kāi)始選擇 k 個(gè) “20”，保證其右端點(diǎn)盡量小，再?gòu)挠疫呴_(kāi)始選擇 k 個(gè) “20”，保證其左端點(diǎn)盡量大，然后依次匹配即可
所以可以二分答案，然后貪心去 check 就好了
代碼： ?
//#pragma GCC optimize(2)
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
using namespace std;typedef long long LL;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const int N=1e6+100;char s[N];int n;vector<int>node1,node2;bool check(int mid)
{if(min(node1.size(),node2.size())<mid*2)return false;for(int i=0;i<mid;i++)if(node1[i]>node2[mid-i-1])return false;return true;
}void init()
{node1.clear(),node2.clear();int pos=1;for(int i=1;i<=n;i++)if(s[i]=='0'){while(pos<i&&s[pos]!='2')pos++;if(pos==i)continue;pos++;node1.push_back(i); }pos=n;for(int i=n;i>=1;i--)if(s[i]=='2'){while(pos>i&&s[pos]!='0')pos--;if(pos==i)continue;pos--;node2.push_back(i);}
}int main()
{
#ifndef ONLINE_JUDGE
// freopen("data.in.txt","r",stdin);
// freopen("data.out.txt","w",stdout);
#endif
// ios::sync_with_stdio(false);while(scanf("%d",&n)!=EOF){scanf("%s",s+1);init();int l=0,r=inf,ans=-1;while(l<=r){int mid=l+r>>1;if(check(mid)){ans=mid;l=mid+1;}elser=mid-1;}printf("%d\n",ans);}return 0;
}
