CCPC-Wannafly Winter Camp Day8 (Div2, onsite)

Replay

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Dup4：

• 厕所是个换换脑子的好地方?
• 要读题啊，不要别人不做，自己就跟着不做啊

 X：

• 读题很重要啊！什么时候才能读对题 不演队友啊 D题看错题， 直到最后一小时才看懂
• 很多时候要看榜单做题

A：Aqours

Solved.

考虑一个点的子树下面有多少个叶子节点汇聚，

那么这个时候就可以更新某些叶子节点的答案

并且把距离该点最近的叶子节点返回上去继续做这一步操作

再正着  考虑除子树外 的离当前点最近的叶子节点的距离，用于更新子树中叶子结点的答案

相当于树形dp起手式

但是这里$n很大，不能DFS$

但是又注意到$i <= j  有 fa[i] < fa[j]$

相当于$bfs序列是1;2;cdots n$

可以直接两次遍历来做这件事情

#include <bits/stdc++.h>
using namespace std;

#define N 3000010
#define pii pair <int, int>
int n, fa[N], ans[N], d[N];   
vector <pii> G[N];
pii g[N];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) G[i].clear();
        memset(ans, -1, sizeof ans);
        memset(d, 0, sizeof d);
        fa[1] = 0; 
        for (int i = 2; i <= n; ++i)
        {
            scanf("%d", fa + i);
            ++d[fa[i]];
        }    
        for (int v = n; v >= 1; --v)
        {
            int u = fa[v];
            if (!d[v]) 
                G[u].push_back(pii(v, 1)); 
            else
            {
                sort(G[v].begin(), G[v].end());
                pii tmp = *G[v].begin();
                ++tmp.second;
                G[u].push_back(tmp);
                int Min = (*G[v].begin()).second;
                for (int i = 1, len = G[v].size(); i < len; ++i)
                {
                    pii it = G[v][i];
                    ans[it.first] = Min + it.second;
                }
            }
        }
        for (int v = 1; v <= n; ++v)
        {
            g[v] = pii(1e9, 1e9);
            int u = fa[v];
            if (u)
            {
                g[v] = g[u];
                ++g[v].second;
                pii tmp = g[v];
                if (tmp.first != -1) for (int i = 0, len = G[v].size(); i < len; ++i) 
                {
                    pii it = G[v][i];
                    if (it.first > tmp.first)    
                        ans[it.first] = min(ans[it.first], it.second + tmp.second);  
                }
            }
            if (!G[v].empty())
            {
                pii it = *G[v].begin();
                if (it.second < g[v].second)
                    g[v] = it;
            }
        }
        for (int i = 1; i <= n; ++i) if (!d[i]) printf("%d %d
", i, ans[i]); 
    }
    return 0;
}

B：玖凛两开花

Upsolved.

我们可以发现，如果答案是$x，那么小于等于x的点的都要匹配到大于等于x的点$

那么我们把$小于x的点放在左边，大于等于x的点放在右边，二分匹配即可$

这样就有一个二分的做法

但实际上一步一步枚举上去也是可行的，$枚举到x + 1的时候，拆掉x + 1原有的匹配关系，再匹配一次即可$

#include <bits/stdc++.h>
using namespace std;

#define N 10010
int n, m, l, r;
vector <int> G[N]; 

int linker[N];
bool used[N];
bool DFS(int u)
{
    for (auto v : G[u]) if (v >= l && v <= r && !used[v])
    {
        used[v] = true;
        if (linker[v] == -1 || DFS(linker[v]))
        {
            linker[v] = u;
            return true; 
        }
    }
    return false;
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = 1; i <= n; ++i) G[i].clear();
        for (int i = 1, u, v; i <= m; ++i) 
        {
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        int res = 0;
        memset(linker, -1, sizeof linker); 
        for (int i = 0; i < n - 1; ++i) 
        {
            l = i + 1, r = n - 1; 
            if (linker[i] != -1)
            {
                memset(used, 0, sizeof used);
                if (!DFS(linker[i])) break;
                linker[i] = -1;
            }
            memset(used, 0, sizeof used);
            if (DFS(i)) ++res;
            else break;
        }
        printf("%d
", res);
    }
    return 0;
}

D：吉良吉影的奇妙计划

Solved.

n只有20

暴力打表

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn = 110;
const ll MOD = 998244353;


int n;
ll ans = 0;
int arr[maxn];
int tmp[maxn];

bool judge()
{
    for(int i = 1; i <= 2 * n; ++i) tmp[i] = arr[i];
    int flag = 1;
    while(flag)
    {
        flag = 0;
        for(int i = 1; i + 1 <= 2 * n; ++i)
        {
            if((tmp[i] == 1 && tmp[i + 1] == 0) || (tmp[i] == 0 && tmp[i + 1] == 1)) 
            {
                tmp[i] = tmp[i + 1] = -1;
                flag = 1;
                break;
            }
        }
    }
    for(int i = 1; i + 3 <= 2 * n; ++i)
    {
        if(tmp[i] + tmp[i + 1] + tmp[i + 2] + tmp[i + 3] == -4) return true;
    }
    return false;
}

void DFS(int cnt, int cnt1, int cnt0)
{
    if(cnt == 2 * n + 1)
    {
        ans = (ans + 1) % MOD;
        return ;
    }
    if(cnt1)
    {
        if(cnt == 1 || (cnt > 1 && arr[cnt - 1] != 0))
        {
            arr[cnt] = 1;
            DFS(cnt + 1, cnt1 - 1, cnt0);
        }
    }
    if(cnt0)
    {
        if(cnt == 1 || (cnt > 1 && arr[cnt - 1] != 1))
        {
            arr[cnt] = 0;
            DFS(cnt + 1, cnt1, cnt0 - 1);
        }
    }
    if(cnt1 > 0 && cnt0 > 0)
    {
        if((cnt + 1 <= 2 * n && arr[cnt - 1] != -1))
        {
            arr[cnt] = -1;
            arr[cnt + 1] = -1;
            DFS(cnt + 2, cnt1 - 1, cnt0 - 1);
            arr[cnt] = -2;
            arr[cnt + 1] = -2;
        }
    }
}

int main()
{
    while(~scanf("%d", &n))
    {
        ans = 0;
        DFS(1, n, n);
        cout << ans << endl;
    }
    return 0;
}

E：Souls-like Game

Solved.

阅读理解题

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 10010
const ll MOD = (ll)998244353;
int n, m, p[N][3][3];

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j < 3; ++j)
                for (int k = 0; k < 3; ++k)
                    scanf("%d", &p[i][j][k]), p[i][j][k] %= MOD;
        }
        for (int i = 1, op, l, r; i <= m; ++i)
        {
            scanf("%d%d%d", &op, &l, &r);
            if (op == 1) 
            {
                int pp[3][3]; 
                for (int j = 0; j < 3; ++j) 
                    for (int k = 0; k < 3; ++k) 
                        scanf("%d", &pp[j][k]), pp[j][k] %= MOD;
                for (int j = l; j <= r; ++j)
                    for (int k = 0; k < 3; ++k)
                        for (int o = 0; o < 3; ++o)
                            p[j][k][o] = pp[k][o];
            }
            else
            {
                ll sum[2][3] = {0, 0, 0};
                for (int j = 0; j < 3; ++j) sum[r & 1][j] = 1;
                for (int j = r - 1; j >= l; --j)
                {
                    for (int k = 0; k < 3; ++k) sum[j & 1][k] = 0;
                    for (int k = 0; k < 3; ++k)
                        for (int o = 0; o < 3; ++o)
                            sum[j & 1][k] = (sum[j & 1][k] + p[j][k][o] * sum[(j & 1) ^ 1][o] % MOD) % MOD;
                }
                ll res = 0;
                for (int j = 0; j < 3; ++j) 
                    res = (res + sum[l & 1][j]) % MOD;
                //for (int j = 0; j < 3; ++j) printf("%lld%c", sum[l & 1][j], " 
"[j == 2]);
                printf("%lld
", res);
            }
        }
    }
    return 0;
}

G：穗乃果的考试

Solved.

$考虑到每个(i, j)位置上的1可能被i cdot (n - i + 1) cdot j cdot (m - j + 1)的矩形包含$

$即每个1都会产生i cdot (n - i + 1) cdot j cdot (m - j + 1)的贡献 累加即可$

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MOD = 998244353;
const int maxn = 2e3 + 10;

int n, m;
char mp[maxn][maxn];

int main()
{
    while(~scanf("%d %d", &n, &m))
    {
        ll ans = 0;
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= m; ++j)
            {
                scanf(" %c", &mp[i][j]);
                if(mp[i][j] == '1')
                {
                    ll tmp = i * (n - i + 1) % MOD * j % MOD * (m - j + 1) % MOD;
                    ans = (ans + tmp) % MOD;
                }
            }
        }
        printf("%lld
", ans);
    }
    return 0;
}

I：岸边露伴的人生经验

Upsolved.

将十维的向量拆成二十位的二进制数

可以通过FWT快速的卷积以下，得到异或后相同的对数

为了方便，我们将距离平方

再将异或值转换成距离

这里注意$dis(1, 2) = 1, 而 1 oplus 2 = 3$ 

$dis(0, 2) = 4, 0 oplus 2 = 2$

然后统计即可

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 2500010
const ll MOD = (ll)998244353; 
const ll inv_2 = (ll)499122177;
int n;
ll c[N];
ll a[110];

void FWT(ll x[],int len,int mode)
{
    for(int i=2;i<=len;i<<=1)
    {
        int step=i>>1;
        for(int j=0;j<len;j+=i)
            for(int k=j;k<j+step;k++)
            {
                ll a=x[k],b=x[k+step];
                x[k]=(a+b)%MOD;
                x[k+step]=(a-b+MOD)%MOD;
                if(mode==-1) (x[k] *= inv_2)%=MOD,(x[k+step]*=inv_2)%=MOD; 
            }
    }
}

int get(int x)
{
    int res = 0;
    for (int i = 19; i >= 0; i -= 2)
    {
        int a = (x >> i) & 1, b = (x >> (i - 1)) & 1;
        if (a == 1 && b == 0) res += 4;
        else if (a == 1 && b == 1) res += 1;
        else res += (a << 1) + b;    
    }
    return res;
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        memset(c, 0, sizeof c);
        memset(a, 0, sizeof a); 
        for (int i = 1, x; i <= n; ++i)
        {
            int tmp = 0; 
            for (int j = 1; j <= 10; ++j) 
            {
                scanf("%d", &x);
                if (x == 0) tmp <<= 2;
                else if (x == 1) tmp <<= 2, tmp += 1;
                else if (x == 2) tmp <<= 1, tmp += 1, tmp <<= 1;  
            }
            ++c[tmp];
        }
        FWT(c, 1 << 20, 1);
        for (int i = 0; i <= 1 << 20; ++i) c[i] = (c[i] * c[i]) % MOD; 
        FWT(c, 1 << 20, -1);     
        for (int i = 0; i <= 1 << 20; ++i)
            a[get(i)] += c[i];  
        ll res = 0;
        for (int i = 0; i <= 40; ++i) res = (res + a[i] * a[i] % MOD) % MOD; 
        printf("%lld
", res);
    }
    return 0;
}