CF1223F. Stack Exterminable Arrays

Solution

奇怪的套路增加了。

大概就是$[l,r]$能匹配完可以轉化為$[1,l?1]$匹配后的狀態和$[1,r]$匹配后的狀態相同，因此$hash$判斷即可。

Code

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <ctime> #include <cassert> #include <string.h> //#include <unordered_set> //#include <unordered_map> //#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; } template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int,int> PR; typedef vector<int> VI;const lod eps=1e-11; const lod pi=acos(-1); const int oo=1<<30; const ll loo=1ll<<62; const int mods=1e9+9; const int MAXN=1000005; const int INF=0x3f3f3f3f;//1061109567 /*--------------------------------------------------------------------*/ inline int read() {int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f; } ull h[MAXN]; map<ull,int> Map; int a[MAXN],f[MAXN],stk[MAXN],SZ=1e9+7; signed main() {int Case=read();while (Case--){ll ans=0;int n=read(),top=0;for (int i=1;i<=n;i++) a[i]=read(),f[i]=0;Map.clear(),Map[0]=n+1,f[n+1]=0;for (int i=n;i>=1;i--){if (top&&stk[top]==a[i]) top--;else stk[++top]=a[i],h[top]=h[top-1]*SZ+a[i];if (Map[h[top]]) f[i]=f[Map[h[top]]]+1;Map[h[top]]=i;ans+=f[i];}printf("%lld\n",ans);}return 0; }

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