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?

X:

  • 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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