『模拟赛』冲刺CSP联训模拟1

合集 - 比赛记录(77)

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Rank

我的我要爆了

A. 几何

上来思路就假了，不知道，样例全过，本来就算假也能拿点，结果绑包了，妈的。

正解 dp，设 $f_{i,j,k}$ 表示串 $s$ 匹配到 $i$ 位，模式串 $x$ 拼接至 $j$ 位，$y$ 拼接至 $k$ 位是否可行，滚动数组优化，复杂度 $\mathcal{O}\mathcal{(}\mathcal{|}\mathcal{s}\mathcal{|}\mathcal{|}\mathcal{x}\mathcal{|}\mathcal{|}\mathcal{y}\mathcal{|}\mathcal{)}$，不太能过，位运算优化一下就能过了，不过数据水，把 string 改成 char 然后改掉大量的取模换成三目运算符也能过。

代码是卡过去的，不放了。

B. 分析

树形 dp。

赛时想的找直径假了就不细说了，又因为绑包所以一分没有。

设计状态 $f_{i,0/1,0/1}$ 表示对于节点 $i$，其度数为奇（1）偶（0）且其子树（不含该点）内有（1）没有（0）度数为奇的点。

考虑转移，可以将其想象为合并两棵子树。对于任意一条边，我们都可以通过操作 A 添加一条重边，即在二者间连两条边，此时其与其子节点的度数奇偶性不变，转移对象是 $f_{u,j_u,k_u|j_v|k_v}$，理解就是合并后子树变成了原先的子树和子节点的子树和子节点，它们的奇偶性共同决定了转移的目标。

再考虑不加边的情况，此时我们考虑如何使用操作 B。一个结论：我们需要的 B 操作次数为 $\frac{\mathrm{度}\mathrm{数}\mathrm{为}\mathrm{奇}\mathrm{的}\mathrm{点}\mathrm{个}\mathrm{数}}{2}-1$。我们此时只在两点间连一条边，奇偶性相反，所以额外花费有：

$$zc=\{\begin{array}{l}-B{j}_{u}and{j}_v\\ B!j_{u}and!j_v\\ 0else\end{array}$$

转移对象同上，只是此时当前节点度数与子节点度数的奇偶性会取反，异或一下就行。

额外的，$f_{i,1,0}$ 这种情况是不存在的。因为一条边会给一棵子树内带来两次度数的增加，如果当前节点的度数为奇，其子树内一定至少有一个节点度数为奇。当然它对答案没影响，求简便的话按上面写就行，想抢最优解可以考虑减少这种情况然后循环展开去做。

点击查看代码

#include<bits/stdc++.h>
#define fo(x,y,z) for(register int (x)=(y);(x)<=(z);(x)++)
#define fu(x,y,z) for(register int (x)=(y);(x)>=(z);(x)--)
using namespace std;
typedef long long ll;
#define lx ll
inline lx qr()
{
	char ch=getchar();lx x=0,f=1;
	for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+(ch^48);
	return x*f;
}
#undef lx
#define qr qr()
#define pii pair<int,int>
#define fi first
#define se second
const int Ratio=0;
const int N=5e5+5;
const int mod=998244353;
int n,A,B;
int hh[N],to[N<<1],ne[N<<1],cnt;
ll f[N][2][2],g[2][2];
namespace Wisadel
{
    inline void Wadd(int u,int v){to[++cnt]=v,ne[cnt]=hh[u],hh[u]=cnt;}
    inline void Wdfs(int u,int fa)
    {
        f[u][0][0]=0;
        f[u][0][1]=f[u][1][0]=f[u][1][1]=1e18;
        for(register int i=hh[u];i!=-1;i=ne[i])
        {
            int v=to[i];
            if(v==fa) continue;
            Wdfs(v,u);
            fo(j,0,1) fo(k,0,1) g[j][k]=f[u][j][k],f[u][j][k]=1e18;
            fo(j1,0,1) fo(j2,0,1) fo(k1,0,1) fo(k2,0,1)
            {
                f[u][j1][j2|k1|k2]=min(f[u][j1][j2|k1|k2],g[j1][j2]+f[v][k1][k2]+A);
                ll zc=(j1&&k1)?-B:(!j1&&!k1)?B:0;
                f[u][j1^1][j2|(k1^1)|k2]=min(f[u][j1^1][j2|(k1^1)|k2],g[j1][j2]+f[v][k1][k2]+zc);
            }
        }
    }
    short main()
    {
        freopen("analyse.in","r",stdin),freopen("analyse.out","w",stdout);
        n=qr,A=qr,B=qr;
        fill(hh+1,hh+1+n,-1);
        fo(i,1,n-1)
        {
            int a=qr,b=qr;
            Wadd(a,b),Wadd(b,a);
        }
        Wdfs(1,0);
        printf("%lld\n",min({f[1][0][0],f[1][1][1]-B,f[1][0][1]-B}));
        return Ratio;
    }
}
int main(){return Wisadel::main();}

C. 代数

die 数

神秘期望题，不过还是被 5k 等大神琢磨出来了，jijidawang 还给出了第二类斯特林数的 $\mathcal{O}\mathcal{(}\mathcal{nk}\mathcal{)}$ 做法，膜拜！

D. 组合

有点像小时候表演过的魔术：你内心选好一个 1~100 之内的数，我拿出几张卡片，你需要回答在卡片上有没有你所想的数字，全部问完后就能知道你选的数是什么。其实原理就是选不同的数回答时的组合不同，也就是这道题的简单版，其实魔术中还有能根据第一个数的特殊快速算出答案带来更好节目效果的设置，但跟这道题就没啥关系了。

我们把每次询问的结果设为 $an{s}_i$，由上基础思想可知能够根据全部的 $an{s}_i$ 可以分析得出选的无序对是多少。根据满分所需 $m\le 26$，$an{s}_i\in [0,1]$，比较好联想到总结果为一个二进制下 26 位数的数，而不同的无序对该数不同。

如果一个数的话就太简单了，有该数为 1 没有为 0，而两个数的话只要询问中有一个有就为 1 否则为 0，发现没有，相当于把数 $a$ 的结果 $x_a$ 与数 $b$ 的结果 $x_b$ 做了位或操作。

这样题意就得到了进一步转化，我们需要找到 1000 个数使得每个无序对 $(a,b)$ 的 $x_a|x_b$ 结果不同。搜了一下网上的构造方法，发现都没有给出 std 中某些特定值的来源，唯一有可能实现的是模拟退火，有兴趣可以实现一下。主要思路就是随数，然后加入答案数组，条件为与当前答案数组内任意数的位或值未出现过，可以用 bitset 标记。求出这 1000 个数后就很简单了，从 1 到 26 位询问这 1000 个数每一位上的状态（1/0）。

感觉 std 做法很神奇但探究这种构造方法意义不大，也许都是试错法得出来的。

std 正常码风

1000 个数输出在文件里。

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
int N=20;
int a[1015];
bitset<1073741824> bk;
int main(){
    int cur=0;
    vector<int> vec;
    for(int i=0;i<(1<<N);i++){
        if(__builtin_popcount(i)==8) vec.pb(i);
    }
    for(int i=0;i<(1<<N);i++){
        if(__builtin_popcount(i)==9) vec.pb(i);
    }
    for(int i=1;i<=1010;i++){
        bool fl2=0;
        for(auto t:vec){
            a[i]=t;
            bool fl=0;
            for(int j=1;j<=i;j++){
                if(bk[a[i]|a[j]]||(i!=j&&i<=996&&__builtin_popcount(a[i]|a[j])<11)){
                    fl=1;
                    for(int k=1;k<j;k++) bk[a[i]|a[k]]=0;
                    break;
                }
                bk[a[i]|a[j]]=1;
            }
            if(fl) continue;
            fl2=1;
            break;
        }
        if(!fl2){
            N++;
            vec.clear();
            for(int j=0;j<(1<<N);j++){
                if(__builtin_popcount(j)==8) vec.pb(j);
            }
            if(N>=24){
                for(int j=0;j<(1<<N);j++){
                    if(__builtin_popcount(j)==9) vec.pb(j);
                }
            }
            if(N==26){
                for(int j=0;j<(1<<N);j++){
                    if(__builtin_popcount(j)!=8&&__builtin_popcount(j)!=9) vec.pb(j);
                }
            }
            i--;
            continue;
        }
        cout<<i<<' '<<N<<endl;
        if(N>26){
            cur=i-1;
            break;
        }
    }
    freopen("122.out","w",stdout);
    for(int i=1;i<=1000;i++) cout<<a[i]<<',';
    cout<<endl;
    return 0;
}

提交答案题，正解代码没意义。

---

末

绑包是最似蚂的行为！

还有 ZROI 这么喜欢放树形 dp 题吗，两天了，明天再有就。。。

题目质量确实很高啊，这次打的错解不算暴力，所以相当于没挂（。

欠的越来越多了，希望早日把那个smjb线段树调出来。

---

完结撒花~

Vibe

Wine out on a Wednesday
Glass full 'cause she thirsty
She said don't drink till it's Thursday
But she'll come out on a workday
When all work and no play could just make Jane go crazy
And it's no fun to worry goin' home leaving early
Then she run it up back it up when she know you feelin' the work
Turn it up just a touch givin' love so good that it hurts
Till she f**k it up leanin up even though she still on the up
Leave you shook turn and looks make you never want to leave
'Cause whether Monday Tuesday Wednesday
You know you could still feel her vibe
Or on a Thursday Friday Saturday
You know she could go one more time
And even if they stop the beat or it's time to leave
You know she could go one last time
'Cause whether her place your place
You know some way you could still feel her vibe
You could still you could still
You could still you could still
You could still feel her vibe
You could still you could still
You could still you could still
You could still feel her vibe
And even if they stop the beat or it's time to leave
You know she could go one last time
'Cause whether her place your place
You know some way you could still feel her vibe
She feelin' caught in the waves
The way she movin' her waist
And you're watchin' that body
The rhythm she rockin' no one could complain
There's a certain kind of grace it takes
To dance like nobody's there
No she don't need no one because the song's still on
'Cause whether Monday Tuesday Wednesday
You know you could still feel her vibe
Or on a Thursday Friday Saturday
You know she could go one more time
And even if they stop the beat or it's time to leave
You know she could go one last time
'Cause whether her place your place
You know some way you could still feel her vibe
You could still you could still
You could still you could still
You could still feel her vibe
You could still you could still
You could still you could still
You could still feel her vibe
And even if they stop the beat or it's time to leave
You know she could go one last time
'Cause whether her place your place
You know some way you could still feel her vibe
Monday Tuesday Monday Tuesday
Or on a Thursday Friday Thursday Friday
Even if they stop the beat if they stop the beat
'Cause whether her place your place her place your place
Vibe
You could still you could still
You could still you could still
You could still feel her vibe
You could still you could still
You could still you could still
You could still feel her vibe
And even if they stop the beat or it's time to leave
You know she could go one last time
'Cause whether her place your place
You know some way you could still feel her vibe