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模板（三）
2022-07-26 09:20:00 【Alex Su (*^▽^*)】
树链剖分
#include <bits/stdc++.h>
#define l(k) (k << 1)
#define r(k) (k << 1 | 1)
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int N = 1e5 + 10;
int t[N << 2], lazy[N << 2], a[N], mod, root, n, m;
int d[N], siz[N], fa[N], son[N], top[N], id[N], rk[N], cnt;
vector<int> g[N];
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return 1LL * a * b % mod; }
void up(int k) { t[k] = add(t[l(k)], t[r(k)]); }
void update(int k, int l, int r, int x)
{
lazy[k] = add(lazy[k], x);
t[k] = add(t[k], mul(r - l + 1, x));
}
void down(int k, int l, int r)
{
if(!lazy[k]) return;
int m = l + r >> 1;
update(l(k), l, m, lazy[k]);
update(r(k), m + 1, r, lazy[k]);
lazy[k] = 0;
}
void build(int k, int l, int r)
{
if(l == r)
{
t[k] = rk[l];
return;
}
int m = l + r >> 1;
build(l(k), l, m);
build(r(k), m + 1, r);
up(k);
}
void change(int k, int l, int r, int L, int R, int x)
{
if(L <= l && r <= R)
{
update(k, l, r, x);
return;
}
down(k, l, r);
int m = l + r >> 1;
if(L <= m) change(l(k), l, m, L, R, x);
if(R > m) change(r(k), m + 1, r, L, R, x);
up(k);
}
int query(int k, int l, int r, int L, int R)
{
if(L <= l && r <= R) return t[k];
down(k, l, r);
int m = l + r >> 1, res = 0;
if(L <= m) res = add(res, query(l(k), l, m, L, R));
if(R > m) res = add(res, query(r(k), m + 1, r, L, R));
return res;
}
void dfs1(int u, int father) //第一次dfs预处理，处理出深度，父亲，大小
{
d[u] = d[father] + 1;
fa[u] = father;
siz[u] = 1;
for(int v: g[u])
{
if(v == father) continue;
dfs1(v, u);
siz[u] += siz[v];
if(siz[son[u]] < siz[v]) son[u] = v;
}
}
void dfs2(int u, int fa, int t) //第二次dfs进行剖分，处理出dfs序，每条链的顶端，以及dfs序对应的权值
{
top[u] = t;
id[u] = ++cnt;
rk[cnt] = a[u];
if(siz[u] == 1) return;
dfs2(son[u], u, t);
for(int v: g[u])
{
if(v == fa || v == son[u]) continue;
dfs2(v, u, v);
}
}
void add(int u, int v, int x)
{
while(top[u] != top[v]) //先以链为整体来考虑，不在同一条链的时候，让链顶端较深的点向上跳
{
if(d[top[u]] < d[top[v]]) swap(u, v);
change(1, 1, n, id[top[u]], id[u], x); //跳的过程重处理这条链上的点的信息
u = fa[top[u]];
}
if(d[u] < d[v]) swap(u, v); //在同一条链时，继续处理
change(1, 1, n, id[v], id[u], x); //从深度小的到深度深的，dfs序从小到大
}
int ask(int u, int v)
{
int res = 0;
while(top[u] != top[v])
{
if(d[top[u]] < d[top[v]]) swap(u, v); //注意是比较top的深度，不是u,v的深度
res = add(res, query(1, 1, n, id[top[u]], id[u]));
u = fa[top[u]];
}
if(d[u] < d[v]) swap(u, v);
res = add(res, query(1, 1, n, id[v], id[u]));
return res;
}
int main()
{
scanf("%d%d%d%d", &n, &m, &root, &mod);
_for(i, 1, n) scanf("%d", &a[i]);
_for(i, 1, n - 1)
{
int u, v;
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(root, 0);
dfs2(root, 0, root);
build(1, 1, n);
while(m--)
{
int op, x, y, z;
scanf("%d", &op);
if(op == 1)
{
scanf("%d%d%d", &x, &y, &z);
add(x, y, z);
}
else if(op == 2)
{
scanf("%d%d", &x, &y);
printf("%d\n", ask(x, y));
}
else if(op == 3)
{
scanf("%d%d", &x, &z);
change(1, 1, n, id[x], id[x] + siz[x] - 1, z); //用size数组推出dfs序的右端点
}
else
{
scanf("%d", &x);
printf("%d\n", query(1, 1, n, id[x], id[x] + siz[x] - 1));
}
}
return 0;
}树上启发式合并
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int son[N], siz[N], c[N], cnt[N], mx, n;
vector<int> g[N];
ll sum, ans[N];
void dfs1(int u, int fa)
{
siz[u] = 1;
for(int v: g[u])
{
if(v == fa) continue;
dfs1(v, u);
siz[u] += siz[v];
if(siz[son[u]] < siz[v]) son[u] = v;
}
}
void clear(int u, int fa)
{
cnt[c[u]]--;
for(int v: g[u])
{
if(v == fa) continue;
clear(v, u);
}
}
void add(int u)
{
cnt[c[u]]++;
if(cnt[c[u]] > mx)
{
mx = cnt[c[u]];
sum = c[u];
}
else if(cnt[c[u]] == mx) sum += c[u];
}
void insert(int u, int fa)
{
add(u);
for(int v: g[u])
{
if(v == fa) continue;
insert(v, u);
}
}
void dfs2(int u, int fa)
{
for(int v: g[u])
{
if(v == fa || v == son[u]) continue;
dfs2(v, u);
mx = sum = 0;
clear(v, u);
}
if(son[u]) dfs2(son[u], u);
for(int v: g[u])
{
if(v == fa || v == son[u]) continue;
insert(v, u);
}
add(u);
ans[u] = sum;
}
int main()
{
scanf("%d", &n);
_for(i, 1, n) scanf("%d", &c[i]);
_for(i, 1, n - 1)
{
int u, v;
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(1, 0);
dfs2(1, 0);
_for(i, 1, n) printf("%lld ", ans[i]);
return 0;
}线性基
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int M = 62;
ll d[M + 10];
bool zero;
void add(ll x)
{
for(int i = M; i >= 0; i--)
if(x & (1LL << i)) //注意这里写1LL防止爆int
{
if(d[i]) x ^= d[i];
else { d[i] = x; return; } //这里是return不是break
}
zero = true; //注意0要特判
}
ll check(ll x) //判断x是否可以用线性基的元素表示 和插入类似
{
for(int i = M; i >= 0; i--)
if(x & (1LL << i))
{
if(d[i]) x ^= d[i];
else return false;
}
return true;
}
ll qmax()
{
ll res = 0;
for(int i = M; i >= 0; i--) //从高到低位贪心 本质就是尽量让高位为1 这样写简洁
res = max(res, res ^ d[i]);
return res;
}
ll qmin() //如果没有0的话 最小的d[i]就是最小值
{
if(zero) return 0;
_for(i, 0, M)
if(d[i])
return d[i];
}
ll kth(int k) //查询所有异或和中第k小的数
{
if(zero) k--;
if(!k) return 0;
_for(i, 0, M) //先将线性基处理一下 注意处理后的线性基仍然可以照常使用
_for(j, 0, i - 1)
if(d[i] & (1LL << j))
d[i] ^= d[j];
ll res = 0;
_for(i, 0, M)
if(d[i])
{
if(k & 1) res ^= d[i];
k >>= 1;
}
return res;
}
int main()
{
int n; scanf("%d", &n);
_for(i, 1, n)
{
ll x; scanf("%lld", &x);
add(x);
}
printf("%lld\n", qmax());
return 0;
}NTT
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int N = 4e6 + 10, mod = 998244353, G = 3, Gi = 332748118; //NTT是FFT的优化，速度更快，避免精度误差
//同时NTT可以取模，模数看题目，这里以998244353为例
int n, m, limit = 1, l, r[N]; //N开数据范围的四倍
ll a[N], b[N], c[N], ans[N];
ll binpow(ll a, ll b)
{
ll res = 1;
for(; b; b >>= 1)
{
if(b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
void NTT(ll *A, int type)
{
rep(i, 0, limit)
if(i < r[i])
swap(A[i], A[r[i]]);
for(int mid = 1; mid < limit; mid <<= 1)
{
ll Wn = binpow(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
for(int j = 0; j < limit; j += (mid << 1))
{
ll w = 1;
for(int k = 0; k < mid; k++, w = w * Wn % mod)
{
int x = A[j + k], y = w * A[j + k + mid] % mod;
A[j + k] = (x + y) % mod;
A[j + k + mid] = (x - y + mod) % mod;
}
}
}
}
void read() //读入n次多项式和m次多项式
{
scanf("%d%d", &n, &m);
_for(i, 0, n) scanf("%d", &a[i]), a[i] = (a[i] + mod) % mod;
_for(i, 0, m) scanf("%d", &b[i]), b[i] = (b[i] + mod) % mod;
}
void solve()
{
while(limit <= n + m) limit <<= 1, l++;
rep(i, 0, limit) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
NTT(a, 1); NTT(b, 1);
rep(i, 0, limit) c[i] = a[i] * b[i] % mod;
NTT(c, -1);
ll inv = binpow(limit, mod - 2);
_for(i, 0, n + m) ans[i] = c[i] * inv % mod;
}
int main()
{
read();
solve();
_for(i, 0, n + m) printf("%d ", ans[i]);
return 0;
}线性求逆
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int N = 5000 + 10;
const int mod = 1e9 + 7;
int inv[N];
int main()
{
inv[0] = inv[1] = 1;
_for(i, 2, 5000) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
return 0;
}带权并查集
#include <cstdio>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int N = 1e5 + 10;
int f[N], r[N], n, m;
int find(int x)
{
if(f[x] == x) return x;
int fa = f[x];
f[x] = find(f[x]);
r[x] = (r[x] + r[fa]) % 2;
return f[x];
}
void Union(int x, int y)
{
int fx = find(x), fy = find(y);
if(fx == fy) return;
f[fx] = fy;
r[fx] = (r[x] + 1 + r[y]) % 2;
}
int main()
{
int T; scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
_for(i, 1, n) f[i] = i, r[i] = 0;
while(m--)
{
char op[5]; int x, y;
scanf("%s%d%d", op, &x, &y);
if(op[0] == 'D') Union(x, y);
else
{
if(find(x) != find(y)) puts("Not sure yet.");
else
{
if(r[x] == r[y]) puts("In the same gang.");
else puts("In different gangs.");
}
}
}
}
return 0;
}线段树合并
#include <bits/stdc++.h>
#define num first
#define sum second
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int ls[N << 6], rs[N << 6], root[N], c[N], cnt, n;
pair<int, ll> t[N << 6];
vector<int> g[N];
ll ans[N];
void up(int k)
{
if(t[ls[k]].num == t[rs[k]].num)
{
t[k] = make_pair(t[ls[k]].num, t[ls[k]].sum + t[rs[k]].sum);
return;
}
t[k] = max(t[ls[k]], t[rs[k]]);
}
void add(int& k, int l, int r, int x)
{
if(!k) k = ++cnt;
if(l == r)
{
t[k].num = 1;
t[k].sum = x;
return;
}
int m = l + r >> 1;
if(x <= m) add(ls[k], l, m, x);
else add(rs[k], m + 1, r, x);
up(k);
}
int merge(int x, int y, int l, int r)
{
if(!x) return y;
if(!y) return x;
if(l == r)
{
t[x].num += t[y].num;
t[x].sum = l;
return x;
}
int m = l + r >> 1;
ls[x] = merge(ls[x], ls[y], l, m);
rs[x] = merge(rs[x], rs[y], m + 1, r);
up(x);
return x;
}
void dfs(int u, int fa)
{
add(root[u], 1, n, c[u]);
for(int v: g[u])
{
if(v == fa) continue;
dfs(v, u);
root[u] = merge(root[u], root[v], 1, n);
}
ans[u] = t[root[u]].sum;
}
int main()
{
scanf("%d", &n);
_for(i, 1, n) scanf("%d", &c[i]);
_for(i, 1, n - 1)
{
int u, v;
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1, 0);
_for(i, 1, n) printf("%lld ", ans[i]);
return 0;
}动态开点线段树
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
int ls[N << 6], rs[N << 6], a[N], cnt, n, m, root; //注意root一定一开始设为0
ll t1[N << 6], t2[N << 6];
void up(int k)
{
t1[k] = t1[ls[k]] + t1[rs[k]];
t2[k] = t2[ls[k]] + t2[rs[k]];
}
void modify(int& k, int l, int r, int x, int p)
{
if(!k) k = ++cnt; //动态开点
if(l == r)
{
t2[k] += p;
t1[k] = t2[k] * x;
return;
}
int m = l + r >> 1;
if(x <= m) modify(ls[k], l, m, x, p);
else modify(rs[k], m + 1, r, x, p);
up(k);
}
ll query1(int k, int l, int r, int L, int R)
{
if(!k) return 0; //询问时询问到空节点返回0
if(L <= l && r <= R) return t1[k];
int m = l + r >> 1; ll res = 0;
if(L <= m) res += query1(ls[k], l, m, L, R);
if(R > m) res += query1(rs[k], m + 1, r, L, R);
return res;
}
int query2(int k, int l, int r, int L, int R)
{
if(!k) return 0;
if(L <= l && r <= R) return t2[k];
int m = l + r >> 1, res = 0;
if(L <= m) res += query2(ls[k], l, m, L, R);
if(R > m) res += query2(rs[k], m + 1, r, L, R);
return res;
}
int main()
{
scanf("%d%d", &n, &m);
modify(root, 0, 1e9, 0, n);
while(m--)
{
char op[5]; int x, y;
scanf("%s%d%d", op, &x, &y);
if(op[0] == 'U')
{
modify(root, 0, 1e9, a[x], -1);
modify(root, 0, 1e9, a[x] = y, 1);
}
else
{
int t = query2(root, 0, 1e9, y, 1e9);
ll sum = query1(root, 0, 1e9, 0, y - 1);
if(sum >= 1LL * (x - t) * y) puts("TAK");
else puts("NIE");
}
}
return 0;
}cdq分治（三维偏序问题）
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int M = 2e5 + 10;
const int N = 1e5 + 10;
struct node
{
int a, b, c, cnt, ans;
}t[N], s[N];
int f[M], k[N], m, n;
int lowbit(int x) { return x & -x; }
void add(int x, int p)
{
for(; x <= m; x += lowbit(x))
f[x] += p;
}
int sum(int x)
{
int res = 0;
for(; x; x -= lowbit(x))
res += f[x];
return res;
}
bool cmp1(node x, node y)
{
if(x.a != y.a) return x.a < y.a;
if(x.b != y.b) return x.b < y.b;
return x.c < y.c;
}
bool cmp2(node x, node y)
{
if(x.b != y.b) return x.b < y.b;
return x.c < y.c;
}
void cdq(int l, int r)
{
if(l == r) return;
int mid = l + r >> 1;
cdq(l, mid); cdq(mid + 1, r);
sort(s + l, s + mid + 1, cmp2);
sort(s + mid + 1, s + r + 1, cmp2);
int i = l, j = mid + 1;
for(; j <= r; j++)
{
while(s[i].b <= s[j].b && i <= mid)
{
add(s[i].c, s[i].cnt);
i++;
}
s[j].ans += sum(s[j].c);
}
_for(t, l, i - 1) add(s[t].c, -s[t].cnt); //注意这里清空不是整个左区间，左区间未必遍历完
}
int main()
{
scanf("%d%d", &n, &m);
_for(i, 1, n) scanf("%d%d%d", &t[i].a, &t[i].b, &t[i].c);
sort(t + 1, t + n + 1, cmp1);
int same = 0, p = 0;
_for(i, 1, n)
{
same++;
if(t[i].a != t[i + 1].a || t[i].b != t[i + 1].b || t[i].c != t[i + 1].c)
{
s[++p] = t[i];
s[p].cnt = same;
same = 0;
}
}
cdq(1, p);
_for(i, 1, p) k[s[i].ans + s[i].cnt - 1] += s[i].cnt;
_for(i, 0, n - 1) printf("%d\n", k[i]);
return 0;
}高斯消元（解线性方程组）
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int N = 15;
double a[N][N], p[N][N];
int n;
void build()
{
scanf("%d", &n);
_for(i, 1, n + 1)
_for(j, 1, n)
scanf("%lf", &p[i][j]);
_for(i, 1, n)
_for(k, 1, n)
{
double x1 = p[i][k], x2 = p[i + 1][k];
a[i][k] = 2 * (x2 - x1);
a[i][n + 1] += x2 * x2 - x1 * x1;
}
}
void Gauss()
{
_for(j, 1, n)
{
int p = j;
while(!a[p][j] && p <= n) p++;
swap(a[p], a[j]);
_for(i, 1, n)
if(i != j)
{
double t = a[i][j] / a[j][j];
_for(k, 1, n + 1) a[i][k] -= t * a[j][k];
}
}
_for(i, 1, n) printf("%.3f ", a[i][n + 1] / a[i][i]);
}
int main()
{
build();
Gauss();
return 0;
}后缀数组
/*
sa[i] 排名为i的后缀的位置
height lcp(sa[i], sa[i - 1]) 即排名为i的后缀与排名为i−1的后缀的最长公共前缀
H[i]：height[rak[i]]，即i号后缀与它前一名的后缀的最长公共前缀
最长公共子串（可重叠） height数组最大值 因为最长公共前缀的两个子串的排名一定相邻
本质不同的字串数目 枚举每一个后缀i 对答案的贡献为 len − sa[i] + 1 − height[i]
两个后缀的最大公共前缀 令x=rank[i],y=rank[j]，x < y,那么lcp(i,j)=min(height[x+1],height[x+2]…height[y])。lcp(i,i)=n-sa[i]。 用ST表或线段树
*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const int N = 1e6 + 10;
int x[N], y[N], c[N], sa[N], rk[N], height[N], wt[30];
int n, m;
char s[N];
void get_SA()
{
_for(i, 1, n) ++c[x[i] = s[i]];
_for(i, 2, m) c[i] += c[i - 1];
for(int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
for(int k = 1; k <= n; k <<= 1)
{
int num = 0;
_for(i, n - k + 1, n) y[++num] = i;
_for(i, 1, n) if(sa[i] > k) y[++num] = sa[i] - k;
_for(i, 1, m) c[i] = 0;
_for(i, 1, n) ++c[x[i]];
_for(i, 2, m) c[i] += c[i - 1];
for(int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i], y[i] = 0;
swap(x, y);
x[sa[1]] = 1; num = 1;
_for(i, 2, n)
x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? num : ++num;
if(num == n) break;
m = num;
}
}
void get_height()
{
int k = 0;
_for(i, 1, n) rk[sa[i]] = i;
_for(i, 1, n)
{
if(rk[i] == 1) continue;
if(k) k--;
int j = sa[rk[i] - 1];
while(j + k <= n && i + k <= n && s[i + k] == s[j + k]) k++;
height[rk[i]] = k;
}
}
int main()
{
scanf("%s", s + 1);
n = strlen(s + 1);
m = 122; //m表示字符个数 ascll('z')=122
//桶的范围是1~m
get_SA();
_for(i, 1, n) printf("%d ", sa[i]);
return 0;
}凸包
#include<bits/stdc++.h>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
const double eps = 1e-8;
const int N = 1e5 + 10;
struct point
{
double x, y;
point(double x = 0, double y = 0) : x(x), y(y) {}
}st[N], p[N];
int n, top;
point operator - (point a, point b) { return {a.x - b.x, a.y - b.y}; }
double operator ^ (point a, point b) { return a.x * b.y - a.y * b.x; } //叉积数学上返回向量，程序这里返回的是值
double dis(point a, point b) { return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2)); }
bool cmp(point a, point b) //横坐标第一关键字，纵坐标第二关键字
{
if(fabs(a.x - b.x) < eps)
return a.y < b.y;
return a.x < b.x;
}
int main()
{
scanf("%d", &n);
_for(i, 1, n) scanf("%lf%lf", &p[i].x, &p[i].y);
sort(p + 1, p + n + 1, cmp);
st[++top] = p[1]; //先加入第一个点，第一个点一定是凸包上的点
_for(i, 2, n) //求下半凸包
{
while(top >= 2 && ((st[top] - st[top - 1]) ^ (p[i] - st[top - 1])) < 0) top--; //这里有一个一条边上多个点的情况，也就是相等的情况
st[++top] = p[i]; //但是这道题不影响结果
} //如果要去掉的话就改成<=0 还要写eps
for(int i = n - 1; i >= 1; i--) //求上半凸包，直接复制，改一下循环顺序就好
{
while(top >= 2 && ((st[top] - st[top - 1]) ^ (p[i] - st[top - 1])) < 0) top--;
st[++top] = p[i];
}
double ans = 0;
_for(i, 1, top - 1) //最后栈里面首尾都是起点，所以可以直接求距离
ans += dis(st[i], st[i + 1]);
printf("%.2lf\n", ans);
return 0;
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