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2026C. Action Figures

思路讲解

总体思路就是一个贪心,想办法省最多的钱。首先,这一天他没有去,那这钱他是省不了了,因为后面的天他去,肯定省后面的多的钱,前面的天也没法买他,因此,我们把这个人加到que0里(实际是一个栈,这种我应该用deque,代码里是vector)。

然后可以省钱的,必须要给他找个搭档,先在que0里找,再在rem(A[i]==’1’的i的set)里找,找小的(通过删除rem.begin()实现)。

AC代码

https://codeforces.com/contest/2026/submission/309701489

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// Problem: C. Action Figures
// Contest: Codeforces - Educational Codeforces Round 171 (Rated for Div. 2)
// URL: https://codeforces.com/problemset/problem/2026/C
// Memory Limit: 512 MB
// Time Limit: 2500 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(4e5)+10,INF=static_cast<ll>(5e18)+3;

ll N,M,T;
char A[MAXN];

inline void solve(){
cin>>N;
ll ans=(1+N)*N/2;
vector<ll> seq0;
set<ll> rem;
for(int i=1;i<=N;++i){
cin>>A[i];
if(A[i]=='0') seq0.push_back(i);
if(A[i]=='1') rem.insert(i);
}
ll idx=0;
// for(int i=N;i>=1;--i){
// if(i<=idx) break;
// if(A[i]=='1' && vis[i]){
//
// ans-=i;
// ++idx;
// }
// }
// #ifdef LOCAL
// cerr << "OK "<<T << '\n';
// #endif
while(!rem.empty()){
set<ll>::iterator it=prev(rem.end());
while(!seq0.empty() && seq0.back()>*it ){
seq0.pop_back();
}
// #ifdef LOCAL
// cerr << seq0.back() << '\n';
// #endif
if(!seq0.empty()){
seq0.pop_back();
ans-=*it;
}else{
if(rem.size()>=2){
rem.erase(rem.begin());
ans-=*it;
// #ifdef LOCAL
// cerr << ans <<" "<< *it << '\n';
// #endif
}
}
rem.erase(it);
}
cout<<ans<<"\n";

}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();

}
return 0;
}
/*
AC
https://codeforces.com/contest/2026/submission/309701489
*/

心路历程(WA,TLE,MLE……)