CCPC-Wannafly Winter Camp Day5 (Div2, onsite)

Replay:

Dup4:

时间复杂度算不对？一点点思路不经过验证就激动的要死？浪费自己一个小时还浪费别人一个小时？
对1e3不敏感？ 1e3 * 1e3是多少？模拟建边跑dp不写非要写个大模拟？
看到数据结构就高兴的要死？没细想？没发现性质？

日常语文差，导致计算几何死都写不对读题要细致啊！
感觉状态还可以?只是计算几何写太久了，人都懵了

A：Cactus Draw

Solved.

按照BFS序以及深度排

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = 1e4 + ;

 struct Edge{

     int to, nxt;

     Edge(){}

     Edge(int to, int nxt) :to(to), nxt(nxt){}

 }edge[maxn << ];

 struct node{

     int x, y;

     node(){}

     node(int x, int y):x(x), y(y){}

 }ans[maxn];

 int n, m;

 int head[maxn], tot;

 int vis[maxn];

 int level[maxn];

 void Init()

 {

     tot = ;

     memset(vis, , sizeof vis);

     memset(level, , sizeof level);

     memset(head, -, sizeof head);

 }

 void addedge(int u,int v)

 {

     edge[tot] = Edge(v, head[u]); head[u] = tot++;

     edge[tot] = Edge(u, head[v]); head[v] = tot++;

 }

 void BFS(int root)

 {

     queue<int>q;

     q.push(root);

     vis[root] = ;

     ans[root] = node(vis[root], ++level[vis[root]]);

     while(!q.empty())

     {

         int u = q.front();

         q.pop();

         for(int i = head[u]; ~i; i = edge[i].nxt)

         {

             int v = edge[i].to;

             if(!vis[v])

             {

                 vis[v] = vis[u] + ;

                 ans[v] = node(vis[v], ++level[vis[v]]);

                 q.push(v);

             }

         }

     }

 }

 int main()

 {

     while(~scanf("%d %d", &n, &m))

     {

         Init();

         for(int i = , u, v; i <= m; ++i)

         {

             scanf("%d %d", &u, &v);

             addedge(u, v);

         }

         BFS();

         for(int i=  ; i <= n; ++i)

         {

             printf("%d %d\n", ans[i].x, ans[i].y);

         }

     }

     return ;

 }

C：Division

Solved.

每次取最大进行操作，堆维护

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define N 100010

 int n, k;

 int main()

 {

     while (scanf("%d%d", &n, &k) != EOF)

     {

         priority_queue <int> pq;

         ll res = ;

         for (int i = , a; i <= n; ++i)

         {

             scanf("%d", &a);

             pq.push(a);

         }

         for (int i = ; i <= k; ++i)

         {

             int top = pq.top(); pq.pop();

             pq.push(top / );

         }

         while (!pq.empty())

         {

             res += pq.top();

             pq.pop();

         }

         printf("%lld\n", res);

     }

     return ;

 }

D：doppelblock

unsolved.

搜索。

增加剪枝：当剩余的数字小于x之间的数字时，回溯掉即可。

还有一条剪枝可以先处理x的位置在填数字（未写）

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const ll MOD = 1e9 + ;

 const int maxn = 1e2 + ;

 int n;

 int sum = ;

 int r[maxn], c[maxn];

 int left_r[maxn], left_c[maxn];

 int mid_r[maxn], mid_c[maxn];

 int right_r[maxn], right_c[maxn];

 int cnt_r[maxn], cnt_c[maxn];

 int vis_r[maxn][maxn], vis_c[maxn][maxn];

 char mp[maxn][maxn];

 void Init()

 {

     memset(left_r, , sizeof left_r);

     memset(left_c, , sizeof left_c);

     memset(mid_r, , sizeof mid_r);

     memset(mid_c, , sizeof mid_c);

     memset(right_r, , sizeof right_r);

     memset(right_c, , sizeof right_c);

     memset(cnt_r, , sizeof cnt_r);

     memset(cnt_c, , sizeof cnt_c);

     memset(vis_r, , sizeof vis_r);

     memset(vis_c, , sizeof vis_c);

 }

 bool DFS(int x, int y)

 {

     if (x == n && y == n + ) return true;

     if (y == n + )

     {

         if (cnt_r[x] != ) return false;

         else return DFS(x + , );

     }

     if (cnt_r[x] ==  && cnt_c[y] == )

     {

         mp[x][y] = 'X';

         cnt_r[x]++;

         cnt_c[y]++;

         if (DFS(x, y + )) return true;

         cnt_r[x]--;

         cnt_c[y]--;

     }

     if (cnt_r[x] ==  && (cnt_c[y] ==  && mid_c[y] == c[y]))

     {

         mp[x][y] = 'X';

         cnt_r[x]++;

         cnt_c[y]++;

         if (DFS(x, y + )) return true;

         cnt_r[x]--;

         cnt_c[y]--;

     }

     if (cnt_c[y] ==  && (cnt_r[x] ==  && mid_r[x] == r[x]))

     {

         mp[x][y] = 'X';

         cnt_r[x]++;

         cnt_c[y]++;

         if (DFS(x, y + )) return true;

         cnt_r[x]--;

         cnt_c[y]--;

     }

     if ((cnt_r[x] ==  && mid_r[x] == r[x]) && (cnt_c[y] ==  && mid_c[y] == c[y]))

     {

         mp[x][y] = 'X';

         cnt_r[x]++;

         cnt_c[y]++;

         if (DFS(x, y + )) return true;

         cnt_r[x]--;

         cnt_c[y]--;

     }

     for (int i = ; i <= n - ; ++i)

     {

         if (vis_r[x][i] || vis_c[y][i]) continue;

         if (cnt_r[x] == )

         {

             if (sum - (left_r[x] + i) < r[x]) continue;

         }

         else if (cnt_r[x] == )

         {

             if (mid_r[x] + i > r[x]) continue;

         }

         if (cnt_c[y] == )

         {

             if (sum - (left_c[y] + i) < c[y]) continue;

         }

         else if (cnt_c[y] == )

         {

             if (mid_c[y] + i > c[y]) continue;

         }

         if (cnt_r[x] == ) left_r[x] += i;

         else if (cnt_r[x] == ) mid_r[x] += i;

         else if (cnt_r[x] == ) right_r[x] += i;

         if (cnt_c[y] == ) left_c[y] += i;

         else if (cnt_c[y] == ) mid_c[y] += i;

         else if (cnt_c[y] == ) right_c[y] += i;

         vis_r[x][i]++;

         vis_c[y][i]++;

         mp[x][y] = i + '';

         if (DFS(x, y + )) return true;

         if (cnt_r[x] == ) left_r[x] -= i;

         else if (cnt_r[x] == ) mid_r[x] -= i;

         else if (cnt_r[x] == ) right_r[x] -= i;

         if (cnt_c[y] == ) left_c[y] -= i;

         else if (cnt_c[y] == ) mid_c[y] -= i;

         else if (cnt_c[y] == ) right_c[y] -= i;

         vis_r[x][i]--;

         vis_c[y][i]--;

     }

     return false;

 }

 void RUN()

 {

     int t;

     int flag = ;

     scanf("%d", &t);

     while (t--)

     {

         if (flag++) printf("\n");

         Init();

         scanf("%d", &n);

         for (int i = ; i <= n; ++i) scanf("%d", r + i);

         for (int i = ; i <= n; ++i) scanf("%d", c + i);

         sum = (n - ) * (n - ) / ;

         DFS(, );

         for (int i = ; i <= n; ++i)

         {

             for (int j = ; j <= n; ++j)

             {

                 printf("%c", mp[i][j]);

             }

             puts("");

         }

     }

 }

 int main()

 {

 #ifdef LOCAL_JUDGE

     freopen("Text.txt", "r", stdin);

 #endif // LOCAL_JUDGE

     RUN();

 #ifdef LOCAL_JUDGE

     fclose(stdin);

 #endif // LOCAL_JUDGE

     return ;

 }

E：Fast Kronecker Transform

Upsolved.

将同样的数放在一起，如果同样的数字小于$10000，直接暴力$

否则做NTT

$因为模数是998244353，可以直接做，做FFT可能有精度问题$

$F(n) = \sum f(t) \cdot g(n - t)$

$f(t) = t 当 a_t = x$

$g(t) = t 当 b_t = x$

 #include <bits/stdc++.h>

 using namespace std;

 #define db long double

 #define ll long long

 #define N 400010

 #define S 10010

 const ll MOD = (ll);

 int n, m, a[N], b[N], c[N];

 vector <int> l[N], r[N];

 ll ans[N];

 void Hash()

 {

     c[] = ;

     for (int i = ; i <= n; ++i) c[++c[]] = a[i];

     for (int i = ; i <= m; ++i) c[++c[]] = b[i];

     sort(c + , c +  + c[]);

     c[] = unique(c + , c +  + c[]) - c - ;

     for (int i = ; i <= n; ++i) a[i] = lower_bound(c + , c +  + c[], a[i]) - c;

     for (int i = ; i <= m; ++i) b[i] = lower_bound(c + , c +  + c[], b[i]) - c;

 }

 ll qmod(ll base, ll n)

 {

     ll res = ;

     while (n)

     {

         if (n & ) res = (res * base) % MOD;

         base = base * base % MOD;

         n >>= ;

     }

     return res;

 }

 int x1[N], x2[N];

 void ntt(int *a, int len, int f)

 {

     int i, j = , t, k;

     for (int i = ; i < len - ; ++i)

     {

         for (t = len; j ^= t >>= , ~j & t;);

         if (i < j) swap(a[i], a[j]);

     }

     for (int i = ; i < len; i <<= )

     {

         t = i << ;

         int wn = qmod(, (MOD - ) / t);

         for (int j = ; j < len; j += t)

         {

             int w = ;

             for (k = ; k < i; ++k, w = 1ll * w * wn % MOD)

             {

                 int x = a[j + k], y = 1ll * w * a[j + k + i] % MOD;

                 a[j + k] = (x + y) % MOD, a[j + k + i] = (x - y + MOD) % MOD;

             }

         }

     }

     if (f == -)

     {

         reverse(a + , a + len);

         int inv = qmod(len, MOD - );

         for (int i = ; i < len; ++i) a[i] = 1ll * a[i] * inv % MOD;

     }

 }

 int main()

 {

     while (scanf("%d%d", &n, &m) != EOF)

     {

         for (int i = ; i <= n; ++i) scanf("%d", a + i);

         for (int i = ; i <= m; ++i) scanf("%d", b + i); Hash();

         for (int i = ; i <= n; ++i) l[a[i]].push_back(i);

         for (int i = ; i <= m; ++i) r[b[i]].push_back(i);

         int len1 = n + , len2 = m + , len = ;

         while (len < (len1 + len2)) len <<= ;

         memset(ans, , sizeof ans);

         for (int i = ; i <= n + m + ; ++i)

         {

             if (l[i].size() + r[i].size() < S)

             {

                 for (auto u : l[i]) for (auto v : r[i])

                     ans[u + v] = (ans[u + v] + (1ll * u * v) % MOD) % MOD;

             }

             else

             {

                 for (int j = ; j < len; ++j) x1[j] = ;

                 for (int j = ; j < len; ++j) x2[j] = ;

                 for (auto x : l[i]) x1[x] = x;

                 for (auto x : r[i]) x2[x] = x;

                 ntt(x1, len, );

                 ntt(x2, len, );

                 for (int j = ; j < len; ++j)

                     x1[j] = 1ll * x1[j] * x2[j] % MOD;

                 ntt(x1, len, -);

                 for (int j = ; j <= n + m; ++j)

                     ans[j] = (ans[j] + x1[j]) % MOD;

             }

         }

         for (int i = ; i <= n + m; ++i) printf("%lld%c", ans[i] % MOD, " \n"[i == n + m]);

     }

     return ;

 }

F：Kropki

Solved.

习惯性记忆化搜索（实际上是个状压dp）

$dp[S][i]表示S状态下i作为最后一个出现的状态， dp下去即可$

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const ll MOD = 1e9 + ;

 int n;

 ll dp[ << ][];

 char str[];

 ll DFS(int S, int last, int dep)

 {

     if(dep == n)

     {

         return 1ll;

     }

     if(dp[S][last] != -) return dp[S][last];

     ll res = ;

     for(int i = ; i <= n; ++i)

     {

         if(S & ( << (i - ))) continue;

         if(dep)

         {

             if(str[dep] == '')

             {

                 if(i != last *  && i *  != last) continue;

             }

             if(str[dep] == '')

             {

                 if(i == last *  || i *  == last) continue;

             }

         }

         ll tmp = DFS((S | ( << (i - ))), i, dep + );

         res = (res + tmp) % MOD;

     }

     dp[S][last] = res;

     return res;

 }

 int main()

 {

     while(~scanf("%d", &n))

     {

         scanf("%s", str + );

         memset(dp, -, sizeof dp);

         ll ans = DFS(, , );

         printf("%lld\n", ans);

     }

     return ;

 }

H：Nested Tree

Solved.

点数只有$10^6，建边树形DP$

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const ll MOD = (ll)1e9 + ;

 const int maxn = 1e6 + ;

 struct Edge{

     int to, nxt;

     Edge(){}

     Edge(int to, int nxt):to(to), nxt(nxt){}

 }edge[maxn << ];

 int n, m;

 int head[maxn], tot;

 ll son[maxn];

 ll ans;

 void Init()

 {

     ans = ;

     tot = ;

     memset(head, -, sizeof head);

 }

 void addedge(int u,int v)

 {

     edge[tot] = Edge(v, head[u]); head[u] = tot++;

     edge[tot] = Edge(u, head[v]); head[v] = tot++;

 }

 void DFS(int u, int fa)

 {

     son[u] = ;

     for(int i = head[u]; ~i; i = edge[i].nxt)

     {

         int v = edge[i].to;

         if(v == fa) continue;

         DFS(v, u);

         son[u] += son[v];

         ans = (ans + (son[v] * (n - son[v]) % MOD) % MOD) % MOD;

     }

 }

 int main()

 {

     while(~scanf("%d %d", &n, &m))

     {

         Init();

         for(int i = , u, v; i < n; ++i)

         {

             scanf("%d %d", &u, &v);

             for(int j = ; j <= m; ++j)

             {

                 addedge((j - ) * n + u, (j - ) * n + v);

             }

         }

         for(int i = , a, b, u, v; i < m; ++i)

         {

             scanf("%d %d %d %d", &a ,&b, &u, &v);

             addedge((a - ) * n + u, (b - ) * n + v);

         }

         n *= m;

         DFS(, -);

         printf("%lld\n", ans);

     }

     return ;

 }

I：Sorting

Upsolved.

将数分为两类，一类是$<= x，二类是> x $

同一类的数在怎么操作其相对位置都是不变的

那么我们只需要知道前缀区间内有多少个一类数，有多少个二类数

再用前缀和维护同一类数的和即可

$2、3操作用线段树维护即可，用0, 1分别表示一类数$

每次操作相当于将前面连续一段赋值为$0/1 后面连续一段赋值为1/0$

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define N 200010

 int n, q, x, a[N];

 ll sum[][N];

 namespace SEG

 {

     int lazy[N << ], v[N << ];

     void pushdown(int id, int l, int r, int mid)

     {

         if (lazy[id] == -) return;

         lazy[id << ] = lazy[id];

         lazy[id <<  | ] = lazy[id];

         v[id << ] = lazy[id] * (mid - l + );

         v[id <<  | ] = lazy[id] * (r - mid);

         lazy[id] = -;

     }

     void pushup(int id) { v[id] = v[id << ] + v[id <<  | ]; }

     void build(int id, int l, int r)

     {

         lazy[id] = -, v[id] = ;

         if (l == r)

         {

             v[id] = a[l] > x;

             return;

         }

         int mid = (l + r) >> ;

         build(id << , l, mid);

         build(id <<  | , mid + , r);

         pushup(id);

     }

     void update(int id, int l, int r, int ql, int qr, int val)

     {

         if (l >= ql && r <= qr)

         {

             lazy[id] = val;

             v[id] = val * (r - l + );

             return;

         }

         int mid = (l + r) >> ;

         pushdown(id, l, r, mid);

         if (ql <= mid) update(id << , l, mid, ql, qr, val);

         if (qr > mid) update(id <<  | , mid + , r, ql, qr, val);

         pushup(id);

     }

     int query(int id, int l, int r, int ql, int qr)

     {

         if (r < l) return ;

         if (l >= ql && r <= qr) return v[id];

         int mid = (l + r) >> ;

         pushdown(id, l, r, mid);

         int res = ;

         if (ql <= mid) res += query(id << , l, mid, ql, qr);

         if (qr > mid) res += query(id <<  | , mid + , r, ql, qr);

         return res;

     }

 }

 ll que(int r)

 {

     if (r < ) return ;

     int a = SEG::query(, , n, , r);

     int b = r - a;

     //cout << a << " " << b << endl;

     //cout << sum[1][a] << " " << sum[0][b] << endl;

     return (a ? sum[][a] : ) + (b ? sum[][b] : );

 }

 int main()

 {

     while (scanf("%d%d%d", &n, &q, &x) != EOF)

     {

         sum[][] = , sum[][] = ;

         for (int i = ; i <= n; ++i)

         {

             scanf("%d", a + i);

             if (a[i] <= x) sum[][++sum[][]] = a[i];

             else sum[][++sum[][]] = a[i];

         }

         for (int i = ; i <= n; ++i) for (int j = ; j < ; ++j) sum[j][i] += sum[j][i - ];

         SEG::build(, , n);

         for (int qq = , op, l, r; qq <= q; ++qq)

         {

             scanf("%d%d%d", &op, &l, &r);

             if (op == ) printf("%lld\n", que(r) - que(l - ));

             else if (op == )

             {

                 int a = SEG::query(, , n, l, r);

                 int b = (r - l + ) - a;

                 SEG::update(, , n, l, l + b - , );

                 SEG::update(, , n, l + b, r, );

             }

             else

             {

                 int a = SEG::query(, , n, l, r);

                 int b = (r - l + ) - a;

                 SEG::update(, , n, l, l + a - , );

                 SEG::update(, , n, l + a, r, );

             }

         }

     }

     return ;

 }

J：Special Judge

Solved.

$枚举每两条边，判一下即可$

 #include<bits/stdc++.h>

 using namespace std;

 const double eps = 1e-;

 const int maxn = 1e4 + ;

 int sgn(__int128 x)

 {

     if(x == ) return ;

     else return x >  ?  : -;

 }

 struct Point{

     __int128 x, y;

     Point(){}

     Point(__int128 _x, __int128 _y)

     {

         x = _x;

         y = _y;

     }

     bool operator == (const Point &b) const

     {

         return sgn(x - b.x) ==  && sgn(y - b.y) == ;

     }

     bool operator < (const Point  &b) const

     {

         return sgn(x - b.x) ==  ? sgn(y - b.y) : sgn(x - b.x);

     }

     Point operator - (const Point &b) const

     {

         return Point(x - b.x, y - b.y);

     }

     __int128 operator ^ (const Point &b) const

     {

         return x * b.y - y * b.x;

     }

     __int128 operator * (const Point &b) const

     {

         return x * b.x + y * b.y;

     }

 }P[maxn];

 struct Line{

     Point s, e;

     Line(){}

     Line(Point _s, Point _e)

     {

         s = _s;

         e = _e;

     }

     void adjust()

     {

         if(e < s) swap(s, e);

     }

     int segcrossseg(Line v)

     {

         int d1 = sgn((e - s) ^ (v.s - s));

         int d2 = sgn((e - s) ^ (v.e - s));

         int d3 = sgn((v.e - v.s) ^ (s - v.s));

         int d4 = sgn((v.e - v.s) ^ (e - v.s));

         if((d1 ^ d2) == - && (d3 ^ d4) == -) return ;

         return (d1 ==  && sgn((v.s - s) * (v.s - e)) <= )

                 || (d2 ==  && sgn((v.e - s) * (v.e - e)) <= )

                 || (d3 ==  && sgn((s - v.s) * (s - v.e)) <= )

                 || (d4 ==  && sgn((e - v.s) * (e - v.e)) <= );

     }

     bool pointtoseg(Point p)

     {

         return sgn((p - s) ^ (e - s)) ==  && sgn((p - s) * (p - e)) <= ;

     }

 }L[maxn];

 int n, m;

 int u[maxn], v[maxn];

 int main()

 {

     while(~scanf("%d %d", &n, &m))

     {

         for(int i = ; i <= m; ++i) scanf("%d %d", u + i, v + i);

         for(int i = ; i <= n; ++i)

         {

             int x, y;

             scanf("%d %d", &x ,&y);

             P[i] = Point(x, y);

         }

         for(int i = ; i <= m; ++i)

         {

                L[i] = Line(P[u[i]], P[v[i]]);

             L[i].adjust();

         }

         int ans = ;

         for(int i = ; i <= m; ++i) for(int j = i + ; j <= m; ++j)

         {

             if(L[i].segcrossseg(L[j]) == ) ans++;

             else if(L[i].segcrossseg(L[j]) == )

             {

                 if(u[i] == u[j])

                 {

                     if(!(L[i].pointtoseg(P[v[j]]) || (L[j].pointtoseg(P[v[i]])))) continue;

                 }

                 if(u[i] == v[j])

                 {

                     if(!(L[i].pointtoseg(P[u[j]]) || (L[j].pointtoseg(P[v[i]])))) continue;

                 }

                 if(v[i] == u[j])

                 {

                     if(!(L[i].pointtoseg(P[v[j]]) || (L[j].pointtoseg(P[u[i]])))) continue;

                 }

                 if(v[i] == v[j])

                 {

                     if(!(L[i].pointtoseg(P[u[j]]) || (L[j].pointtoseg(P[u[i]])))) continue;

                 }

                 ans++;

             }

         }

         printf("%d\n", ans);

     }

     return ;

 }

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