CF diary
似乎不只有cf
1334E. Divisor Paths
\(\texttt{Difficulty:2200}\)
题意
一个正整数 \(D(1\le D\le10^{15})\) 。由 \(D\) 建出一张无向图，每个节点都是 \(D\) 的约数， 存在边 \((u,v)(u>v)\) 当且仅当存在素数 \(p\) 使得 \(v\cdot p=u\) ，边权为 \(u,v\) 的约数个数的差，现在有 \(q（1\le q\le10^5)\) 次询问，每次询问 \(u\) 到 \(v\) 的最短路数量，模 \(998244353\) 。
题解
设 \(u>v\) ，如果 \(v\) 是 \(u\) 的约数，那么从 \(v\) 不断乘上一些素数走到 \(u\) 即是最短路，我们对 \(\frac{u}{v}\) 分解质因数求一个多重集的排列即可。如果 \(v\) 不是 \(u\) 的约数，可以发现最短路一定会经过 \(gcd(u,v)\) 中转，于是分别求出 \(u\) 到 \(gcd(u,v)\) 和 \(v\) 到 \(gcd(u,v)\) 的最短路，相乘即可。由于所有数都是 \(D\) 的约数，他们的质因数分解可以在对 \(D\) 质因数分解后，逐个尝试 \(D\) 的质因数来求得。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 200010;
LL D, Q, fact[110], invfact[110];
LL qpow(LL a, LL x, LL m)
{
LL ans = 1;
while (x)
{
if (x & 1)
ans = ans * a % m;
x >>= 1;
a = a * a % m;
}
return ans % m;
}
void fact_init(LL n, LL m)
{
fact[0] = fact[1] = 1;
for (LL i = 2; i <= n; i++)
fact[i] = fact[i - 1] * i % m;
invfact[n] = qpow(fact[n], m - 2, m);;
for (LL i = n; i > 0; i--)
invfact[i - 1] = invfact[i] * i % m;
}
map<LL, LL>mp;
LL gcd(LL a, LL b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
LL calc(LL u, LL v)
{
LL nd = u / v;
vector<LL>tmp;
int all = 0;
for (auto& [x, y] : mp)
{
int cnt = 0;
while (nd % x == 0)
nd /= x, cnt++;
if (cnt)
tmp.push_back(cnt);
all += cnt;
}
LL ans = fact[all];
for (auto& c : tmp)
ans = ans * invfact[c] % mod;
return ans;
}
void solve()
{
fact_init(100, mod);
LL x = D;
for (LL i = 2; i * i <= x; i++)
{
while (x % i == 0)
mp[i]++, x /= i;
}
if (x != 1)
mp[x]++;
LL u, v;
while (Q--)
{
cin >> u >> v;
if (u == v)
{
cout << 1 << endl;
continue;
}
if (u < v)
swap(u, v);
LL k = gcd(u, v);
if (k == v)
cout << calc(u, v) << endl;
else
cout << calc(u, k) * calc(v, k) % mod << endl;
}
}
int main()
{
IOS;
cin >> D >> Q;
solve();
return 0;
}
1322B. Present
\(\texttt{Difficulty:2100}\)
题意
一个长为 \(n(1\le n\le4\cdot10^5)\) 的序列 \(a(1\le a_i\le 10^7) \) ，求对其每个二元组求和后全部异或起来的结果。
题解
考虑每一位对于答案的贡献，两个数相加时如果第 \(k\) 位为 \(1\) ，那么其 \(0\sim k\) 位组成的数相加的值 \(\in[2^k,2^{k+1})\cup[2^{k+1}+2^k,+\infty)\) 。我们对于每个 \(k\) ，将原序列取模取出 \(0\sim k\) 位的数字排序后双指针即可得到该位上 \(1\) 的个数，根据奇偶性来判断最后答案上该位的值即可。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 400010;
int N, A[maxn], B[maxn];
void solve()
{
int ans = 0;
for (int i = 0; i <= 25; i++)
{
for (int j = 1; j <= N; j++)
B[j] = A[j] % (1 << (i + 1));
sort(B + 1, B + N + 1);
int cnt = 0;
for (int l = 1, r = N; l <= N; l++)
{
while (r > l && B[l] + B[r] >= (1 << i))
r--;
if (r <= l)
r = l;
cnt += N - r;
}
for (int l = 1, r = N; l <= N; l++)
{
while (r > l && B[l] + B[r] >= (1 << (i + 1)))
r--;
if (r <= l)
r = l;
cnt -= N - r;
}
for (int l = 1, r = N; l <= N; l++)
{
while (r > l && B[l] + B[r] >= (1 << (i + 1)) + (1 << i))
r--;
if (r <= l)
r = l;
cnt += N - r;
}
if (cnt & 1)
ans |= 1 << i;
}
cout << ans << endl;
}
int main()
{
IOS;
cin >> N;
for (int i = 1; i <= N; i++)
cin >> A[i];
solve();
return 0;
}
1495C. Garden of the Sun
\(\texttt{Difficulty:2300}\)
题意
一个 \(n\cdot m(1\le n\cdot m\le250000)\) 的网格，每个点.代表空地，X 代表太阳花，所有初始太阳花的格子之间没有公共边和点，要求种上任意数量的太阳花，使得所有太阳花四联通并且没有环。
题解
考虑由于初始的太阳花之间两两没有公共边和点，我们考虑每隔两行就将改行种满，这样太阳花形成了若干连通块，之后考虑将它们连接起来，中间间隔的两行找一个不会产生环的列，在该列上两个位置都种上花，对于边界处多出两行的情况，每列最后一行的位置上只要有花，就在倒数第二行该列也种上花即可，特判一下只有一列的请况。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<LL, LL>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define pb push_back
//#define int LL
//#define lc p*2
//#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-8;
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 510;
int T, N, M;
char A[maxn][maxn];
void solve()
{
if (M == 1)
{
for (int i = 1; i <= N; i++)
cout << 'X' << endl;
return;
}
for (int i = 1; i <= N; i += 3)
{
for (int j = 1; j <= M; j++)
A[i][j] = 'X';
}
for (int i = 2; i + 1 <= N; i += 3)
{
if (i + 1 == N)
{
for (int j = 1; j <= M; j++)
{
if (A[N][j] == 'X')
A[N - 1][j] = A[N][j] = 'X';
}
}
else
{
for (int j = 1; j <= M; j++)
{
if (j == 1)
{
if (A[i][j + 1] == '.' && A[i + 1][j + 1] == '.')
{
A[i][j] = A[i + 1][j] = 'X';
break;
}
}
else if (j == N)
{
if (A[i][j - 1] == '.' && A[i + 1][j - 1] == '.')
{
A[i][j] = A[i + 1][j] = 'X';
break;
}
}
else
{
if (A[i][j + 1] == '.' && A[i + 1][j + 1] == '.' && A[i][j - 1] == '.' && A[i + 1][j - 1] == '.')
{
A[i][j] = A[i + 1][j] = 'X';
break;
}
}
}
}
}
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= M; j++)
cout << A[i][j];
cout << endl;
}
}
int main()
{
IOS;
cin >> T;
while (T--)
{
cin >> N >> M;
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= M; j++)
cin >> A[i][j];
}
solve();
}
return 0;
}
1059D. Nature Reserve
\(\texttt{Difficulty:2200}\)
题意
\(n(1\le n\le 10^5)\) 个点，求包含所有点且与 \(x\) 轴相切的最小圆的半径，无解输出-1。
题解
首先如果 \(x\) 轴两侧都有点或者其轴上有多于一个点则无解，对于有解的情况，为了便于处理，所有点都在 \(y\) 轴负方向时我们将它们都对称到正方向来。我们显然可以二分半径，对于每个半径 \(r\) ，圆心在 \(y=r\) 上。考虑如何 \(\texttt{check}\) ，对于每个点，圆心能包含它的一段横坐标是连续的，我们可以对每个点求出这段区间，然后取交集，如果不为空，则说明可行，否则不行，复杂度 \(O(nlogn)\) 。实现时注意精度，式子尽可能化简后再计算。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<LL, LL>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define pb push_back
//#define int LL
//#define lc p*2
//#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-8;
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 100010;
int N;
PII P[maxn];
int cmp(double x)
{
if (fabs(x) < eps)
return 0;
if (x < 0)
return -1;
return 1;
}
pair<LD, LD> calc(LD x1, LD y1, LD y)
{
return make_pair(x1 - sqrt(2.0 * y1 * y - y1 * y1), x1 + sqrt(2.0 * y1 * y - y1 * y1));
}
bool check(LD y)
{
LD l = -1e8, r = 1e8;
for (int i = 1; i <= N; i++)
{
pair<LD, LD>seg = calc(P[i].first, P[i].second, y);
if (cmp(seg.second - l) < 0 || cmp(seg.first - r) > 0)
return false;
l = max(l, seg.first), r = min(r, seg.second);
}
return true;
}
void solve()
{
bool ne = false, po = false, zr = false;
for (int i = 1; i <= N; i++)
{
if (zr && P[i].second == 0)
{
cout << -1 << endl;
return;
}
if (P[i].second == 0)
zr = true;
else if (P[i].second > 0)
po = true;
else
ne = true;
if (ne && po)
{
cout << -1 << endl;
return;
}
}
for (int i = 1; i <= N; i++)
P[i].second = abs(P[i].second);
LD lo = 0, hi = 1e18;
for (int i = 1; i <= 500; i++)
{
LD mid = (lo + hi) / 2;
if (check(mid))
hi = mid;
else
lo = mid;
}
cout << setiosflags(ios::fixed) << setprecision(12) << hi << endl;
}
int main()
{
IOS;
cin >> N;
for (int i = 1; i <= N; i++)
cin >> P[i].first >> P[i].second;
solve();
return 0;
}
P6146 [USACO20FEB]Help Yourself G
题意
\(n(1\le n\le 10^5)\) 个线段，其一个子集的复杂度为集合中线段取并后的连续段数量，求所有子集的复杂度之和。
题解
按左端点排序，记 \(f_i\) 为之在前 \(i\) 个线段中进行选取的复杂度之和，对于 \(f_{i+1}\) ，如果不考虑额外的贡献，第 \(i+1\) 条线段可选或不选，这部分为 \(f_i\cdot2\) ，对于额外的贡献，只有在已选的集合中没有右端点大于自己左端点的线段时才会产生，这部分记录一下这种线段的数量为 \(cntr\) ，则额外的贡献为 \(2^{cntr}\) ，递推即可。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 100010;
int N;
PII S[maxn];
LL cntr[maxn * 2], f[maxn];
LL qpow(LL a, LL x)
{
LL ans = 1;
while (x)
{
if (x & 1)
ans = ans * a % MOD;
a = a * a % MOD, x >>= 1;
}
return ans;
}
void solve()
{
sort(S + 1, S + N + 1);
for (int i = 1; i <= N; i++)
cntr[S[i].second]++;
for (int i = 1; i <= N * 2; i++)
cntr[i] += cntr[i - 1];
for (int i = 1; i <= N; i++)
f[i] = (f[i - 1] * 2LL % MOD + qpow(2LL, cntr[S[i].first - 1])) % MOD;
cout << f[N] << endl;
}
int main()
{
IOS;
cin >> N;
for (int i = 1; i <= N; i++)
cin >> S[i].first >> S[i].second;
solve();
return 0;
}
1009F. Dominant Indices
\(\texttt{Difficulty:2300}\)
题意
一颗根为 \(1\) 的 \(n（1\le n\le10^6)\) 的树，对于每个点 \(u\) ，求使 \(d(u,x)\) 最大的最小 \(x\) ， \(d(u,x)\) 为 \(u\) 子树中距离 \(u\) 为 \(x\) 点的数量。
题解
题目要求查询所有子树的信息，考虑 \(\texttt{dsu on tree}\) 。维护当前子树中每个深度的节点个数，同时维护最大值以及答案，这部分在累加时很好更新，清除轻子树贡献时也只需将最大值与答案都置为 \(0\) 即可。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 1000010;
int N, tot = 0, hson[maxn], siz[maxn], dep[maxn], dfn[maxn], rnk[maxn], cnt[maxn], ans[maxn], tmp = 0, mx = 0, mxdp = 0;
vector<int>G[maxn];
void add(int v)
{
cnt[dep[v]]++;
if (cnt[dep[v]] > mx)
tmp = dep[v], mx = cnt[dep[v]];
else if (cnt[dep[v]] == mx)
tmp = min(tmp, dep[v]);
}
void add_edge(int from, int to)
{
G[from].push_back(to);
G[to].push_back(from);
}
void dfs1(int v, int p, int d)
{
siz[v] = 1, dfn[v] = ++tot, rnk[tot] = v, dep[v] = d, mxdp = max(mxdp, d);
for (auto& to : G[v])
{
if (to == p)
continue;
dfs1(to, v, d + 1), siz[v] += siz[to];
if (!hson[v] || siz[to] > siz[hson[v]])
hson[v] = to;
}
}
void dfs2(int v, int p)
{
for (auto& to : G[v])
{
if (to == p || to == hson[v])
continue;
dfs2(to, v), tmp = mx = 0;
for (int i = dfn[to]; i < dfn[to] + siz[to]; i++)
cnt[dep[rnk[i]]]--;
}
if (hson[v])
dfs2(hson[v], v);
for (auto& to : G[v])
{
if (to == p || to == hson[v])
continue;
for (int i = dfn[to]; i < dfn[to] + siz[to]; i++)
add(rnk[i]);
}
add(v), ans[v] = tmp - dep[v];
}
void solve()
{
dfs1(1, 0, 1), dfs2(1, 0);
for (int i = 1; i <= N; i++)
cout << ans[i] << endl;
}
int main()
{
IOS;
cin >> N;
int u, v;
for (int i = 1; i < N; i++)
cin >> u >> v, add_edge(u, v);
solve();
return 0;
}
208E. Dominant Indices
\(\texttt{Difficulty:2100}\)
题意
一个 \(n(1\le n\le10^5)\) 个节点的森林， \(m(1\le m\le10^5)\) 次询问，每次询问 \(v,p\) ，求 \(v\) 有多少 \(p\) 级表兄弟，即有多少节点和 \(v\) 的 \(p\) 级祖先相同。
题解
考虑将询问离线，询问变为在 \(v\) 的 \(p\) 级祖先处询问有多少深度和 \(v\) 相同的点，减一即为答案。 \(p\) 级祖先可以倍增查找，维护子树内各深度节点数量直接 \(\texttt{dsu on tree}\) 维护即可。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 100010;
int N, M, root, tot = 0, dfn[maxn], siz[maxn], rnk[maxn], hson[maxn], dep[maxn], ans[maxn], cnt[maxn], fa[20][maxn];
vector<int>G[maxn];
vector<PII>Q[maxn];
void add(int v, int x)
{
cnt[dep[v]] += x;
}
void add_edge(int from, int to)
{
G[from].push_back(to);
G[to].push_back(from);
}
void dfs1(int v, int p, int d)
{
siz[v] = 1, dfn[v] = ++tot, rnk[tot] = v, dep[v] = d, fa[0][v] = p;
for (auto& to : G[v])
{
if (to == p)
continue;
dfs1(to, v, d + 1), siz[v] += siz[to];
if (!hson[v] || siz[to] > siz[hson[v]])
hson[v] = to;
}
}
void dfs2(int v, int p)
{
for (auto& to : G[v])
{
if (to == p || to == hson[v])
continue;
dfs2(to, v);
for (int i = dfn[to]; i < dfn[to] + siz[to]; i++)
add(rnk[i], -1);
}
if (hson[v])
dfs2(hson[v], v);
for (auto& to : G[v])
{
if (to == p || to == hson[v])
continue;
for (int i = dfn[to]; i < dfn[to] + siz[to]; i++)
add(rnk[i], 1);
}
add(v, 1);
for (auto& [x, y] : Q[v])
ans[y] = cnt[x] - 1;
}
void solve()
{
dfs1(root, 0, 1);
for (int i = 0; i + 1 < 20; i++)
{
for (int v = 1; v <= N; v++)
{
if (fa[i][v] <= 0)
fa[i + 1][v] = 0;
else
fa[i + 1][v] = fa[i][fa[i][v]];
}
}
cin >> M;
int v, p;
for (int i = 1; i <= M; i++)
{
cin >> v >> p;
int k = dep[v];
for (int j = 19; j >= 0; j--)
{
if (!fa[j][v])
continue;
if (dep[fa[j][v]] >= k - p)
v = fa[j][v];
}
if (v != root)
Q[v].push_back(PII(k, i));
}
dfs2(root, 0);
for (int i = 1; i <= M; i++)
cout << ans[i] << ' ';
cout << endl;
}
int main()
{
IOS;
cin >> N, root = N + 1;
int p;
for (int i = 1; i <= N; i++)
{
cin >> p;
if (!p)
add_edge(root, i);
else
add_edge(p, i);
}
solve();
return 0;
}
1721E. Prefix Function Queries
\(\texttt{Difficulty:2200}\)
题意
给一个串 \(s(1\le|s|\le10^6)\) ， \(q(1\le q\le10^5)\) 个串 \(t(1\le|t_i|\le10)\) 。分别求将串 \(|t_i|\) 接到 \(s\) 后面后最后 \(|t_i|\) 个前缀的 \(\texttt{border}\) 长度。
题解
考虑暴力的做法是先对 \(s\) 跑 \(\texttt{kmp}\) ，在依次处理每个 \(t_i\) ，跑剩下部分的 \(\texttt{kmp}\) ，这样的话复杂度为 \(O(q(|s|+|t|))\) 显然不行。 于是有个叫 \(\texttt{kmp}\) 自动机的东西，我们用 \(fail\) 来代表普通 \(\texttt{kmp}\) 的 \(nxt\) ， \(nxt\) 改为代表该自动机上的 \(trans\) 边，下标从 \(1\) 开始。
记 \(nxt_{i,j}\) 为从 \(i\) 开始沿着 \(fail\) 树跳到的最大的一个位置 \(k\) ，有 \(s_{k+1}=j\) 。那么我们可以由此对 \(\texttt{kmp}\) 中求前缀函数的过程进行加速， \(\texttt{kmp}\) 中这一过程即从 \(fail_{i-1}\) 不断沿 \(fail\) 树跳到的最大的一个位置 \(k\) ，有 \(s_{k+1}=s_i\) ，如果没有找到则变为 \(0\) 。之后 \(fail_i=k\) ，根据 \(nxt\) 的定义，可以得到 \(fail_i=nxt_{fail_{i-1},s_i}\) 。那么对于 \(nxt_{i,j}\) 的求解，如果 \(j=s_{i+1}\) ，那么 \(nxt_{i,j}=i\) ，否则 \(nxt_{i,j}=nxt_{fail_{i,j}}\)
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 100010;
const int maxm = 1000020;
int Q, N;
string S, T[maxn];
int fail[maxm], nxt[maxm][26];
void kmp()
{
for (int i = 2, j = 0; i <= N; i++)
{
while (j >= 1 && S[i - 1] != S[j + 1 - 1])
j = fail[j];
if (S[i - 1] == S[j + 1 - 1])
j++;
fail[i] = j;
}
for (int i = 0; i < N; i++)
{
for (int j = 0; j < 26; j++)
nxt[i][j] = nxt[fail[i]][j];
nxt[i][S[i + 1 - 1] - 'a'] = i;
}
}
void solve()
{
N = S.length();
kmp();
for (int i = 0; i < 10; i++)
S += '#';
int A = N + 1;
for (int i = 1; i <= Q; i++)
{
int M = N + T[i].length();
for (int j = 0; j < T[i].length(); j++)
S[N + j] = T[i][j];
for (int j = A, k = fail[N]; j <= M; j++)
{
int x = j - 1;
k = nxt[k][S[j - 1] - 'a'];
if (S[j - 1] == S[k + 1 - 1])
k++;
fail[j] = k, cout << k << ' ';
for (int l = 0; l < 26; l++)
nxt[x][l] = nxt[fail[x]][l];
nxt[x][S[j - 1] - 'a'] = x;
}
cout << endl;
}
}
int main()
{
IOS;
cin >> S >> Q;
for (int i = 1; i <= Q; i++)
cin >> T[i];
solve();
return 0;
}
1139E. Maximize Mex
\(\texttt{Difficulty:2400}\)
题意
\(n(1\le n\le5000)\) 个人， \(m(1\le m\le5000)\) 个社团，每个人有能力值 \(p(0\le p\le n)\) ，以及所属社团 \(c(1\le c\le m)\) 。总共 \(d(1\le d\le n)\) 天，每天开始前第 \(k_i(1\le k_i\le n)\) 会离开。求从每个社团选出 \(1\) 个人（没有就不选）时，团队的最大能力值，其定义为，所有选出来的人的能力值集合的 \(mex\) 。
题解
考虑将社团设为一部分点，能力值设为另外一部分点，询问离线后改为加边，从小到大进行二分图匹配，如果当前点不能匹配，则记录答案，下个询问从当前节点继续匹配即可。
代码
view code
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 5010;
int N, M, P[maxn], C[maxn], D, K[maxn], ans[maxn], match[maxn * 2];
bool flag[maxn], used[maxn * 2];
vector<int>G[maxn * 2];
void add_edge(int from, int to)
{
G[from].push_back(to);
G[to].push_back(from);
}
bool dfs(int v)
{
used[v] = true;
for (auto& to : G[v])
{
int w = match[to];
if (w < 0 || !used[w] && dfs(w))
{
match[v] = to, match[to] = v;
return true;
}
}
return false;
}
void solve()
{
for (int i = 1; i <= N; i++)
{
if (flag[i])
continue;
add_edge(P[i], C[i] + N);
}
int tmp = 0;
mst(match, -1);
for (int i = D, j = 0; i >= 1; i--)
{
for (; j <= N + M; j++)
{
if (match[j] < 0)
{
mst(used, 0);
if (dfs(j))
tmp++;
else
break;
}
}
add_edge(P[K[i]], C[K[i]] + N), ans[i] = tmp;
}
for (int i = 1; i <= D; i++)
cout << ans[i] << endl;
}
int main()
{
IOS;
cin >> N >> M;
for (int i = 1; i <= N; i++)
cin >> P[i];
for (int i = 1; i <= N; i++)
cin >> C[i];
cin >> D;
for (int i = 1; i <= D; i++)
cin >> K[i], flag[K[i]] = true;
solve();
return 0;
}
1721D. Maximum AND
\(\texttt{Difficulty:1800}\)
来点水题
题意
长为 \(n(1\le n\le10^5)\) 的数列 \(a,b(0\le a_i,b_i<2^{30})\) ，记 \(f(a,b)\) 为 & \(_{i=1}^n a_i\oplus b_i\) ，现在你可以任意排列 \(b\) ，求 \(f(a,b)\) 的最大值。
题解
我们按位考虑，对于第 \(k\) 位，如果其能够为 \(1\) ，我们记 \(x_i,y_i\) 分别为 \(a_i,b_i\) 模 \(2^k\) 后的数，那么就必须存在一种排列，使得所有 \(x_i\oplus y_i\) & \(tmp\) 仍然为 \(tmp\) 。由于 \(tmp\) 中为 \(0\) 的位是无所谓的，我们可以先将所有的 \(x_i,y_i\) 与上 \(tmp\) 。接下来考虑如何判断，发现我们对 \(x,y\) 按不同的顺序排序，然后对次排列进行 \(check\) 即可判断了，因为当前所有的数或上 \(tmp\) 后仍为 \(tmp\) ，如果最大的数字想要异或一个数达到 \(tmp\) ，那么只可能去异或当前最小的（因为不去异或最小的那么最小的就不可能找到一个数异或为 \(tmp\)），于是可以在 \(O(nlognloga)\) 解决。
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TP = tuple<int, int, int>;
#define all(x) x.begin(),x.end()
#define mst(x,v) memset(x,v,sizeof(x))
#define mul(x,y) (1ll*(x)*(y)%mod)
#define mk make_pair
//#define int LL
//#define double LD
#define lc p*2
#define rc p*2+1
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#pragma warning(disable : 4996)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const double eps = 1e-10;
const double pi = acos(-1);
const LL MOD = 1000000007;
const LL mod = 998244353;
const int maxn = 100010;
int T, N, A[maxn], B[maxn], X[maxn], Y[maxn];
void solve()
{
int ans = 0;
for (int i = 29; i >= 0; i--)
{
bool flag = true;
int tmp = (ans | (1 << i)) >> i;
for (int j = 1; j <= N; j++)
X[j] = (A[j] >> i) & tmp, Y[j] = (B[j] >> i) & tmp;
sort(X + 1, X + N + 1), sort(Y + 1, Y + N + 1);
for (int j = 1; j <= N; j++)
{
int num = X[j] ^ Y[N - j + 1];
if (num != tmp)
{
flag = false;
break;
}
}
if (flag)
ans |= (1 << i);
}
cout << ans << endl;
}
int main()
{
IOS;
cin >> T;
while (T--)
{
cin >> N;
for (int i = 1; i <= N; i++)
cin >> A[i];
for (int i = 1; i <= N; i++)
cin >> B[i];
solve();
}
return 0;
}
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