[考试总结]noip模拟40

最近真的是爆炸啊。。。

到现在还是有不少没改出来。。。。

所以先写一下 \(T1\) 的题解。。。。

送花

我们移动右端点,之后我们用线段树维护全局最大值。

之后还要记录上次的位置和上上次的位置。

之后每次加和减。



#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define asm(i,x) for(register signed i=head[x];i;i=edge[i].next)
namespace xin_io
{
	#define debug cout<<"debug"<<endl
	#define enum(x) cout<<#x" = "<<x<<endl;
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll; typedef unsigned long long ull;
	class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &x)
	{
		register type s = 0; register int f = 1; register char ch = gc();
		while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
		while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch  xor 48),ch = gc(); return x = s * f,*this;
	}}io;
}
using namespace xin_io; static const int maxn = 5e6+10,inf = 1e9+7,mod = 998244353; const ll llinf = 1e18+7;
#define int long long
namespace xin
{
	int lst1[maxn],lst2[maxn];
	class xin_seg
	{
		private:
			#define ls(fa) (fa << 1)
			#define rs(fa) (fa << 1 | 1)
			inline void up(int fa) {t[fa].s = std::max(t[ls(fa)].s , t[rs(fa)].s);}
			inline void down(int fa)
			{
				if(!t[fa].debt) return ;
				t[ls(fa)].s += t[fa].debt ; t[rs(fa)].s += t[fa].debt;
				t[ls(fa)].debt += t[fa].debt ; t[rs(fa)].debt += t[fa].debt;
				t[fa].debt = 0;
			}
		public:
			class xin_tree{public:int s,debt;}t[maxn];
			void update(int fa,int l,int r,int ql,int qr,int val)
			{
				if(ql <= l and qr >= r)
				{
					t[fa].s += val;
					t[fa].debt += val;
					return ;
				}
				register int mid = l + r >> 1;
				down(fa);
				if(ql <= mid) update(ls(fa),l,mid,ql,qr,val);
				if(qr >  mid) update(rs(fa),mid+1,r,ql,qr,val);
				up(fa);
			}
	}t;
	int n,c[maxn],d[maxn],m;
	int ans = -inf;
	inline short main()
	{
		io >> n >> m;
		try(i,1,n) io >> c[i];
		try(i,1,m) io >> d[i];
		try(i,1,n)
		{
//			cout<<"i = "<<i<<" c[i] = "<<c[i]<<" lst1[c[i]] = "<<lst1[c[i]]<<" lst2[c[i]] = "<<lst2[c[i]]<<endl;
			if(!lst1[c[i]] and !lst2[c[i]]) t.update(1,1,n,1,i,d[c[i]]);
			else if(lst1[c[i]] and !lst2[c[i]]) t.update(1,1,n,lst1[c[i]]+1,i,d[c[i]]),t.update(1,1,n,1,lst1[c[i]],-d[c[i]]);
			else t.update(1,1,n,lst2[c[i]]+1,lst1[c[i]],-d[c[i]]),t.update(1,1,n,lst1[c[i]]+1,i,d[c[i]]);
			ans = std::max(ans,t.t[1].s);
//			enum(t.t[1].s);
			lst2[c[i]] = lst1[c[i]];
			lst1[c[i]] = i;
		}
		cout<<ans<<endl;
		return 0;
	}
}
signed main() {return xin::main();}

posted @ 2021-08-17 19:35  NP2Z  阅读(35)  评论(0编辑  收藏  举报