【2020省选模拟】题解

T1:

傻逼般的把暴力写挂了
考虑最左右的两个对称点一定在$k+1$以内
暴力枚举后即可确定对称中心
双指针判断有多少对对称点满足即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005;
char xxx;
int cnt,n,k,ans;
pii p[N];
inline pii operator +(cs pii &a,cs pii &b){
	return pii(a.fi+b.fi,a.se+b.se);
}
map<pii,int>vt;
inline void calc(pii now){
	if(vt[now])return;
	vt[now]=1;
	int l=1,r=n,cnt=0;
	while(l<=r){
		if(p[l]+p[r]==now)l++,r--;
		else{
			cnt++;
			if(p[l]+p[r]<now)l++;
			else r--;
		}
	}
	if(cnt<=k)ans++;
}
char yyy;
int main(){
	n=read(),k=read();
	if(k>=n)return puts("-1"),0;
	for(int i=1;i<=n;i++)p[i].fi=read(),p[i].se=read();
	sort(p+1,p+n+1);
	for(int i=1;i<=k+1;i++)
	for(int j=1;j<=k+1;j++)
	calc(p[i]+p[n-j+1]);
	cout<<ans<<'
';
	return 0;
}

T2：

先求得$k$次后某一边被吹掉$i$块的概率$f[i]$
显然有一个用前后缀和优化的$n^3dp$
即记$s[l][r]$表示当前行保留第$[l,r]$个块的概率

考虑如果用$s1[r]=sum_ls[l][r],s2[l]=sum_{r}s[l][r]$
发现把贡献拆开可以维护$f,s1,s1*f$等一堆前缀和得到
复杂度$O(n^2)$

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(int x){return (x<0)?x+mod:x;}
cs int N=1505,M=100005;
int fac[M],ifac[M];
inline void init_inv(cs int len=M-5){
	fac[0]=ifac[0]=1;
	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
	ifac[len]=Inv(fac[len]);
	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline int C(int n,int m){return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m]));}
char xx;
int f[N],pf[N],n,m,p,q,k,pw1[M],pw2[M];
int s1[N],s2[N],pre[N],suf[N],spre[N],ssuf[N];
char yy;
int main(){
	n=read(),m=read(),p=read(),q=read();
	p=mul(p,Inv(q)),q=dec(1,p);
	k=read(),pw1[0]=pw2[0]=1;
	init_inv();
	for(int i=1;i<=k;i++)pw1[i]=mul(pw1[i-1],p),pw2[i]=mul(pw2[i-1],q);
	for(int i=0;i<=k;i++)Add(f[min(i,m)],mul(C(k,i),mul(pw1[i],pw2[k-i])));
	pf[0]=f[0];
	for(int i=1;i<=m;i++)pf[i]=add(pf[i-1],f[i]);
	s1[m]=1,s2[1]=1;
	int ss=0;
	for(int i=1;i<=n;i++){
		for(int k=1;k<=m;k++)
			pre[k]=s1[k];
		for(int j=1;j<=m;j++)
			suf[j]=s2[j]; 
		for(int j=1;j<=m;j++)Add(pre[j],pre[j-1]),spre[j]=add(spre[j-1],mul(pre[j],f[j]));
		for(int j=m;j>=1;j--)Add(suf[j],suf[j+1]),ssuf[j]=add(ssuf[j+1],mul(suf[j],f[m-j+1]));
		ss=pre[m];
		for(int k=1;k<=m;k++){
			int ret=mul(ss,pf[k-1]);Dec(ret,mul(suf[k+1],pf[k-1]));Dec(ret,spre[k-1]);
			s1[k]=mul(ret,f[m-k]); 
		}
		for(int k=1;k<=m;k++){
			int now=mul(ss,pf[m-k]);Dec(now,mul(pre[k-1],pf[m-k])),Dec(now,ssuf[k+1]);
			s2[k]=mul(now,f[k-1]);
		}
	}
	int ret=0;
	for(int k=1;k<=m;k++)Add(ret,s1[k]);
	cout<<ret<<'
';
	return 0;
}

T3：

考虑先对第一种边求出最小生成树
从小到大枚举第二种边
找到换上最大的第一种边即可得到用哪种边的分界点
用$LCT$维护即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define pil pair<int,ll>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005;
cs int inf=1e9+7;
namespace Lct{
	cs int N=::N*5;
	int son[N][2],fa[N],mx[N],mxpos[N],val[N],pos[N],rev[N];
	#define lc(u) son[u][0]
	#define rc(u) son[u][1]
	inline void init(int u,int v,int p){
		mxpos[u]=pos[u]=p,val[u]=mx[u]=v;
	}
	inline void pushup(int u){
		mx[u]=val[u],mxpos[u]=pos[u];
		if(lc(u)&&mx[lc(u)]>mx[u])mx[u]=mx[lc(u)],mxpos[u]=mxpos[lc(u)];
		if(rc(u)&&mx[rc(u)]>mx[u])mx[u]=mx[rc(u)],mxpos[u]=mxpos[rc(u)];
	}
	inline void pushnow(int u){
		swap(lc(u),rc(u)),rev[u]^=1;
	}
	inline void pushdown(int u){
		if(!rev[u])return;
		pushnow(lc(u)),pushnow(rc(u)),rev[u]=0;
	}
	inline bool isrc(int u){
		return rc(fa[u])==u;
	}
	inline bool isrt(int u){
		return !fa[u]||(lc(fa[u])!=u&&rc(fa[u])!=u);
	}
	inline void rotate(int v){
		int u=fa[v],z=fa[u];
		int t=isrc(v);
		if(!isrt(u))son[z][isrc(u)]=v;
		fa[v]=z;
		son[u][t]=son[v][t^1];
		fa[son[v][t^1]]=u;
		son[v][t^1]=u,fa[u]=v;
		pushup(u),pushup(v);
	}
	int stk[N],top;
	inline void splay(int u){
		stk[top=1]=u;
		for(int v=u;!isrt(v);v=fa[v])stk[++top]=fa[v];
		for(int i=top;i;i--)pushdown(stk[i]);
		while(!isrt(u)){
			if(!isrt(fa[u]))
			isrc(fa[u])==isrc(u)?rotate(fa[u]):rotate(u);
			rotate(u);
		}
	}
	inline void access(int u){
		for(int v=0;u;v=u,u=fa[u]){
			splay(u);
			rc(u)=v,pushup(u);
		}
	}
	inline void makert(int u){
		access(u),splay(u),pushnow(u);
	}
	inline void link(int u,int v){
		access(u),splay(u),makert(v),fa[v]=u;
	}
	inline void cut(int u,int v){
		makert(u),access(v),splay(v);
		fa[lc(v)]=0,lc(v)=0,pushup(v);
	}
	inline int query(int u,int v){
		makert(u),access(v),splay(v);
		return mxpos[v];
	}
}
int n,a,b,q;
struct edge{
	int u,v,w;
	friend inline bool operator <(cs edge &a,cs edge &b){
		return a.w<b.w;
	}
}e1[N*2],e2[N*2];
int fa[N],cnt;
ll upd[N];
ll s[N];
inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
inline void solve(){
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=a;i++){
		int f1=find(e1[i].u),f2=find(e1[i].v);
		if(f1!=f2){
			fa[f1]=f2,s[0]+=e1[i].w;
			Lct::link(e1[i].u,i+n),Lct::link(i+n,e1[i].v);
		}
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=b;i++){
		int u=e2[i].u,v=e2[i].v,f1=find(u),f2=find(v);
		if(f1!=f2){
			fa[f1]=f2;int pos=Lct::query(u,v);
			upd[++cnt]=e2[i].w-e1[pos].w;
			Lct::cut(e1[pos].u,pos+n),Lct::cut(e1[pos].v,pos+n);
			Lct::link(u,a+n+i),Lct::link(v,a+n+i);
		}
	}
	sort(upd+1,upd+cnt+1);
	for(int i=1;i<=cnt;i++)s[i]=s[i-1]+upd[i];
	while(q--){
		int x=read(),pos=upper_bound(upd+1,upd+cnt+1,x*2)-upd-1;
		cout<<s[pos]+1ll*(n-1-pos*2)*x<<'
';
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),a=read(),b=read(),q=read();
	for(int i=1;i<=a;i++)e1[i].u=read(),e1[i].v=read(),e1[i].w=read();
	for(int i=1;i<=b;i++)e2[i].u=read(),e2[i].v=read(),e2[i].w=read();
	sort(e1+1,e1+a+1),sort(e2+1,e2+b+1);
	for(int i=1;i<=n;i++)Lct::init(i,-inf,0);
	for(int i=1;i<=a;i++)Lct::init(i+n,e1[i].w,i);
	for(int i=1;i<=b;i++)Lct::init(n+a+i,-inf,0);
	solve();return 0;
}