2022NOIP A层联测35

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最后一篇正经博客了.

下发文件（密码为原 accoders 比赛密码）

传送门

弹珠游戏

发现 R、G、B 的个数对于每个人，选的状态一定只有选的最多的、最多和其次的、最多其次与最少的. 求一下方案数，乘一下 n! 即可（就是每个人排列）.

点击查看代码

#include<bits/stdc++.h>
#define ll long long
#define rg register
#define rll rg ll
#define pll pair<ll,ll>
#define maxn 300001
#define mod 998244353
#define put_ putchar(' ')
#define putn putchar('\n')
using namespace std;
inline ll read()
{
	rg bool f=0;rll x=0;rg char ch=getchar();while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=getchar();return f?-x:x;
}
inline void write(rll x) { if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10|'0'); }
ll n,tag[maxn],ans=1;
char s[maxn];
vector<ll> posr,posg,posb;
int main()
{
	freopen("A.in","r",stdin); freopen("A.out","w",stdout);
	n=read(); scanf("%s",s+1);
	for(rll i=1;i<=(n<<1)+n;i++) (s[i]=='R'?posr:s[i]=='G'?posg:posb).emplace_back(i);
	for(rll i=0;i<n;i++) tag[min({ posr[i],posg[i],posb[i] })]=-1,tag[max({ posr[i],posg[i],posb[i] })]=1;
	for(rll i=1,l=0,r=0;i<=(n<<1)+n;i++)
	{
		if(!~tag[i]) l++; else if(!tag[i]) (ans*=l--)%=mod,r++; else (ans*=r--)%=mod;
	}
	for(rll i=1;i<=n;i++) (ans*=i)%=mod; write(ans);
	return 0;
}

晚会

并查集，求最大生成树.

点击查看代码

#include<bits/stdc++.h>
#define ll long long
#define rg register
#define rll rg ll
#define pll pair<ll,ll>
#define maxn 300001
#define put_ putchar(' ')
#define putn putchar('\n')
using namespace std;
inline ll read()
{
	rg bool f=0;rll x=0;rg char ch=getchar();while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=getchar();return f?-x:x;
}
inline void write(rll x) { if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10|'0'); }
struct node
{
	ll a,b,d;
	inline friend bool operator<(rg node a,rg node b) { return a.d>b.d; }
}a[maxn];
ll n,m,f[maxn],sz[maxn],tot,ans;
inline ll find(rll x) { if(x^f[x]) f[x]=find(f[x]); return f[x]; }
int main()
{
	freopen("B.in","r",stdin); freopen("B.out","w",stdout);
	n=read();m=read(); for(rll i=1;i<=m;i++) a[i]=(node){ read(),read(),read() }; sort(a+1,a+m+1);
	for(rll i=1;i<=n;i++) sz[i]=1,f[i]=i;
	for(rll i=1;i<=m;i++)
	{
		if((i^1)&&a[i].d<a[i-1].d)
			for(rll j=i;j<=m;j--) { if(a[i].d^a[j].d) break; if(find(a[j].a)==find(a[j].b)) { puts("-1");return 0; }}
		if(find(a[i].a)^find(a[i].b))
			tot+=sz[find(a[i].a)]*sz[find(a[i].b)],sz[find(a[i].a)]+=sz[find(a[i].b)],f[find(a[i].b)]=find(a[i].a);
		ans+=(a[i].d-a[i+1].d)*tot;
	}
	write(ans+=(n*(n-1)>>1)-tot);
	return 0;
}

优美的字符串

选举

莫队+线段树. 将询问按照左端点所在块分组.

点击查看代码

#include<bits/stdc++.h>
#define ll long long
#define rg register
#define rll rg ll
#define pll pair<ll,ll>
#define maxn 100001
#define put_ putchar(' ')
#define putn putchar('\n')
using namespace std;
inline ll read()
{
	rg bool f=0;rll x=0;rg char ch=getchar();while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=getchar();return f?-x:x;
}
inline void write(rll x) { if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10|'0'); }
ll n,m,q,sq,a[maxn],b[maxn],mx[maxn],st[maxn],ed[maxn],bel[maxn],ans[maxn],t[maxn];
vector<pll> g[maxn];
unordered_map<ll,ll> mp;
struct node
{
	ll l,r,vl,vr,id;
	inline friend bool operator<(rg node a,rg node b) { if(bel[a.l]==bel[b.l]) return a.r<b.r; return bel[a.l]<bel[b.l]; }
}c[maxn];
class block
{
	public:
	ll sq,v[maxn],mx[maxn],st[maxn],ed[maxn],bel[maxn];
	inline void init()
	{
		sq=__builtin_sqrt(m); for(rll i=1;i<=sq;i++) st[i]=m/sq*(i-1)+1,ed[i]=m/sq*i; ed[sq]=m;
		for(rll i=1;i<=sq;i++) for(rll j=st[i];j<=ed[i];j++) bel[j]=i;
	}
	inline void clear() { for(rll i=1;i<=sq;i++) mx[i]=0; for(rll i=1;i<=m;i++) v[i]=0; }
	inline void upd(rll x,rll y) { mx[bel[x]]=max(mx[bel[x]],v[x]+=y); }
	inline ll query(rll l,rll r)
	{
		rll ans=0; if(bel[l]==bel[r]) { for(rll i=l;i<=r;i++) ans=max(ans,v[i]); return ans; }
		for(rll i=l;i<=ed[bel[l]];i++) ans=max(ans,v[i]); for(rll i=st[bel[r]];i<=r;i++) ans=max(ans,v[i]);
		for(rll i=bel[l]+1;i<bel[r];i++) ans=max(ans,mx[i]); return ans;
	}
}s;
inline void upd(rll x,rg bool mod) { s.upd(a[x],mod?b[x]:-b[x]); }
int main()
{
	freopen("D.in","r",stdin); freopen("D.out","w",stdout);
	n=read();m=read();q=read(); sq=__builtin_sqrt(n); for(rll i=1;i<=n;i++) a[i]=read(); for(rll i=1;i<=n;i++) b[i]=read();
	for(rll i=1;i<=sq;i++) st[i]=n/sq*(i-1)+1,ed[i]=n/sq*i; ed[sq]=n; for(rll i=1;i<=sq;i++) for(rll j=st[i];j<=ed[i];j++) bel[j]=i;
	for(rll i=1;i<=q;i++) c[i]=(node){ read(),read(),read(),read(),i }; sort(c+1,c+q+1); s.init();
	for(rll i=1,l,r,x=1;i<=sq;i++)
	{
		s.clear();l=ed[i]+1,r=ed[i];while(x<=q&&bel[c[x].l]<i) x++;
		while(x<=q&&bel[c[x].l]==i)
		{
			if(bel[c[x].l]==bel[c[x].r])
			{
				rll t=0;mp.clear(); for(rll j=c[x].l;j<=c[x].r;j++) { mp[a[j]]+=b[j]; if(a[j]>=c[x].vl&&a[j]<=c[x].vr) t=max(t,mp[a[j]]); }
				ans[c[x].id]=t; x++; continue;
			}
			while(r<c[x].r) upd(++r,1); for(rll j=1;j<=s.sq;j++) t[j]=s.mx[j];
			while(l>c[x].l) upd(--l,1); ans[c[x].id]=s.query(c[x].vl,c[x].vr);
			while(l<=ed[i]) upd(l++,0); for(rll j=1;j<=s.sq;j++) s.mx[j]=t[j]; x++;
		}
	}
	for(rll i=1;i<=q;i++) write(ans[i]),putn;
	return 0;
}