c++算法竞赛常用板子集合（持续更新）

前言

本文主要包含算法竞赛一些常用的板子，码风可能不是太好，还请见谅。

后续会继续补充没有的板子。当然我太菜了有些可能写不出来T^T

稍微有些分类但不多，原谅我QwQ

建议 Ctrl + F 以快速查找板子。

常用板子

树状数组

此处为查询区间和的树状数组。

int bit[500010];
void add(int k, int x) {
	while (k <= n) {
		bit[k] += x;
		k += lowbit(k);
	}
}
int ask(int k) {
	int res = 0;
	while (k) {
		res += bit[k];
		k -= lowbit(k);
	}
	return res;
}

线段树

此处为区间修改区间查询区间和的线段树。

struct SegmentTree {
	ll sum[N << 2], lazy[N << 2];
	int l[N << 2], r[N << 2];
	void update(int rt) {
		sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
	}
	void pushdown(int rt) {
		if (!lazy[rt]) return ;
		sum[rt << 1] += (r[rt << 1] - l[rt << 1] + 1) * lazy[rt], lazy[rt << 1] += lazy[rt];
		sum[rt << 1 | 1] += (r[rt << 1 | 1] - l[rt << 1 | 1] + 1) * lazy[rt], lazy[rt << 1 | 1] += lazy[rt];
		lazy[rt] = 0;
		update(rt);
	}
	void build(int rt, int L, int R) {
		l[rt] = L, r[rt] = R;
		if (L == R) {
			sum[rt] = a[L];
			return ;
		}
		int mid = L + R >> 1;
		build(rt << 1, L, mid), build(rt << 1 | 1, mid + 1, R);
		update(rt);
	}
	void change(int rt, int L, int R, int x) {
		if (L <= l[rt] && r[rt] <= R) {
			sum[rt] += (r[rt] - l[rt] + 1) * x;
			lazy[rt] += x;
			return ;
		}
		pushdown(rt);
		if (L <= r[rt << 1]) change(rt << 1, L, R, x);
		if (l[rt << 1 | 1] <= R) change(rt << 1 | 1, L, R, x);
		update(rt);
	}
	ll query(int rt, int L, int R) {
		if (L <= l[rt] && r[rt] <= R) return sum[rt];
		pushdown(rt);
		ll res = 0;
		if (L <= r[rt << 1]) res += query(rt << 1, L, R);
		if (l[rt << 1 | 1] <= R) res += query(rt << 1 | 1, L, R);
		return res;
	}
} tree;

堆

不是吧真有人手写堆吗

ll q[N], cnt;
void pushup(int id) {
	while (id > 1) {
		if (q[id] >= q[id >> 1]) break;
		swap(q[id], q[id >> 1]);
		id >>= 1;
	}
}
void movedown() {
	int id = 1;
	while (id << 1 <= cnt) {
		if ((id << 1 | 1) <= cnt) {
			if (q[id] < min(q[id << 1], q[id << 1 | 1])) break;;
			if (q[id << 1] < q[id << 1 | 1]) swap(q[id], q[id << 1]), id <<= 1;
			else swap(q[id], q[id << 1 | 1]), id = id << 1 | 1;
		}
		else {
			if (q[id] > q[id << 1]) swap(q[id], q[id << 1]);
			break;
		}
	}
}
void add(ll x) {
	q[++cnt] = x;
	pushup(cnt);
}
void pop() {
	swap(q[1], q[cnt]);
	cnt--;
	movedown();
}

并查集

struct Disjoint_Set {
	int p[N], size[N];
	void build() {
		for (int i = 1; i <= n; i++) p[i] = i, size[i] = 1;
	}
	int root(int x) {
		if (p[x] != x) return p[x] = root(p[x]);
		return x;
	}
	void merge(int x, int y) {
		x = root(x), y = root(y);
		if (size[x] > size[y]) swap(x, y);
		p[x] = y;
		size[y] += size[x];
	}
	bool check(int x, int y) {
		x = root(x), y = root(y);
		return x == y;
	}
} a;

ST表

代码实现查询区间 \([l, r]\) 的区间最大值

for (int i = 1; i <= n; i++) st[0][i] = a[i];
for (int j = 1; j <= lg; j++) {
	for (int i = 1; i <= n - (1 << j) + 1; i++) {
		st[j][i] = max(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
	}
}
int l, r, lg2, len;
for (int i = 1; i <= m; i++) {
	l = read(), r = read();
	lg2 = log2(r - l + 1);
	len = 1 << lg2;
	printf("%d\n", max(st[lg2][l], st[lg2][r - len + 1]));
}

边链表

const int N = 100010;
int last[N], cnt;
struct edge {
	int to, next, w;
} e[N << 1];
void addedge(int x, int y, int w) {
	e[++cnt].to = y;
	e[cnt].next = last[x];
	e[cnt].w = w;
	last[x] = cnt;
}

LCA

此处贴的是 Tarjan法 求LCA。更多方法

struct Disjoint_Set {
	int p[N], size[N];
	void build() {
		for (int i = 1; i <= n; i++) p[i] = i, size[i] = 1;
	}
	int root(int x) {
		if (p[x] != x) return p[x] = root(p[x]);
		return x;
	}
	void merge(int x, int y) {
		x = root(x), y = root(y);
		if (size[x] > size[y]) swap(x, y);
		p[x] = y;
		size[y] += size[x];
	}
	bool check(int x, int y) {
		x = root(x), y = root(y);
		return x == y;
	}
} a;
int last[N], cnt;
struct edge {
	int to, next;
} e[N << 1];
void addedge(int x, int y) {
	e[++cnt].to = y;
	e[cnt].next = last[x];
	last[x] = cnt;
}
struct node {
	int x, y, ans;
} ask[N];
vector <int> g[N];
int p[N];
bool vis[N];
int r[N];
void dfs(int x, int f) {
	p[x] = f;
	for (int i = last[x]; i; i = e[i].next) {
		int v = e[i].to;
		if (v == f) continue;
		vis[v] = 1;
		for (int j : g[v]) {
			int o = ask[j].x;
			if (o == v) o = ask[j].y;
			if (!vis[o]) continue;
			ask[j].ans = r[a.root(o)];
		}
		dfs(v, x);
		a.merge(x, v);
		r[a.root(x)] = x;
	}
}

单源最短路（Dijkstra）

这里是堆优化版呢。笑了有些时候堆优化还没不优化好

void dij(int s) {
	priority_queue <pii, vector<pii>, greater<pii> > q;
	memset(dis, 0x7f7f7f7f, sizeof(dis));
	q.push({0, s});
	dis[s] = 0;
	while (!q.empty()) {
		pii u = q.top(); q.pop();
		int pos = u.second;
		if (vis[pos]) continue;
		vis[pos] = 1;
		for (int j = last[pos]; j; j = e[j].next) {
			int v = e[j].to;
			if (vis[v]) continue;
			if (dis[pos] + e[j].w < dis[v]) dis[v] = dis[pos] + e[j].w, q.push({dis[v], v});
		}
	}

缩点

其中 \(p\) 为缩点后的新点。

int dfn[N], low[N], dcnt;
bool instack[N];
stack <int> s;
int p[N], h[N];
void dfs(int x, int f) {
	instack[x] = 1;
	s.push(x);
	dfn[x] = low[x] = ++dcnt;
	for (int i = last[0][x]; i; i = e[0][i].next) {
		int v = e[0][i].to;
		if (dfn[v]) {
			if (instack[v]) low[x] = min(low[x], dfn[v]);
			continue;
		}
		dfs(v, x);
		low[x] = min(low[x], low[v]);
	}
	if (low[x] >= dfn[x]) {
		p[x] = x, h[x] = a[x], instack[x] = 0;
		while (s.top() != x) {
			p[s.top()] = x;
			h[x] += a[s.top()];
			instack[s.top()] = 0;
			s.pop();
		}
		s.pop();
	}
}

欧拉路径

int st[N], ed[N];
struct edge {
	int u, v;
} e[N << 1];
int rd[N], cd[N];
bool cmp(edge x, edge y) {
	if (x.u != y.u) return x.u < y.u;
	return x.v < y.v;
}
int ans[N << 1], cnt;
void dfs(int x) {
	while (st[x] <= ed[x]) {
		st[x]++;
		dfs(e[st[x] - 1].v);
	}
	ans[++cnt] = x;
}

乘法逆元

fac[0] = fac[1] = 1;
for (int i = 2; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
inv[1] = 1;
for (int i = 2; i <= n; i++) inv[i] = (mod - mod / i) * inv[mod % i] % mod;

快速幂

ll qpow(ll a, ll b) {
	ll res = 1;
	while (b) {
		if (b & 1) res = res * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return res;
}

矩阵快速幂

不是我说这写的是真的丑，凑活着看吧QAQ

struct sq {
	ll x[110][110];
	void build() {
		for (int i = 1; i <= n; i++) x[i][i] = 1;
	}
	void dd() {
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				x[i][j] = 0;
	}
} a, ans;
sq operator *(const sq &x, const sq &y) {
	sq res;
	res.dd();
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				res.x[i][j] = (res.x[i][j] + x.x[i][k] * y.x[k][j] % mod) % mod;
	return res;
}
void qpow(ll x) {
	while (x) {
		if (x & 1) ans = ans * a;
		a = a * a;
		x >>= 1;
	}
}

线性基

\(p\) 数组表示基底，\(x\) 为添加进的数字。

int p[N];
void add(ll x) {
	for (int i = N; i >= 0; i--) {
		if (!(x & (1ll << i))) continue;
		if (p[i]) x ^= p[i];
		else {p[i] = x; return ;}
	}
}

线性筛

int prime[6000010], cnt;
bool isprime[N + 10];
void prim() {
	isprime[0] = isprime[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (!isprime[i]) prime[++cnt] = i;
		for (int j = 1; j <= cnt && i * prime[j] <= n; j++) {
			isprime[i * prime[j]] = 1;
			if (i % prime[j] == 0) break;
		}
	}
}

字符串哈希

int Char(char c) {
	if (c >= '0' && c <= '9') return c - '0' + 1; //0~9: 1~10
	if (c >= 'a' && c <= 'z') return c - 'a' + 11; //a~z: 11~37
	if (c >= 'A' && c <= 'Z') return c - 'A' + 38; //A~Z: 38~65
	return 0;
}
map <ll, int> mp;

cin >> s;
ll x = 0;
for (int i = 0; i < s.size(); i++) x = (x * 100) + Char(s[i]);
mp[x] = 1;

KMP

\(s\) 和 \(t\) 为需要匹配的两个 char 类型数组。

\(border_i\) 表示 \(t\) 长度为 \(i\) 的前缀最长的 \(border\) 长度。

完了border是啥来着？

ls = strlen(s + 1), lt = strlen(t + 1);
int j = 0;
for (int i = 2; i <= lt; i++) {
	while (j >= 1 && t[j + 1] != t[i]) j = border[j];
	if (t[j + 1] == t[i]) j++;
	border[i] = j;
}
int sx = 1, tx = 0;
while (sx <= ls) {
	while (tx >= 1 && s[sx] != t[tx + 1]) tx = border[tx];
	if (t[tx + 1] == s[sx]) tx++;
	if (tx == lt) printf("%d\n", sx - lt + 1);
	sx++;
}

AC自动机

struct Trie {
	int id[27], cnt, fail;
} t[N];
void Build(string &s) {
	int now = 0;
	for (int i = 0; i < s.size(); i++) {
		if (!t[now].id[s[i] - 'a']) t[now].id[s[i] - 'a'] = ++cnt;
		now = t[now].id[s[i] - 'a'];
	}
	t[now].cnt++;
}
void Fail() {
	queue <int> q;
	for (int i = 0; i < 26; i++) {
		int v = t[0].id[i];
		if (v != 0) {
			t[v].fail = 0;
			q.push(v);
		}
	}
	while (!q.empty()) {
		int u = q.front(); q.pop();
		for (int i = 0; i < 26; i++) {
			int v = t[u].id[i];
			if (v != 0) {
				t[v].fail = t[t[u].fail].id[i];
				q.push(v);
			}
			else t[u].id[i] = t[t[u].fail].id[i];
		}
	}
}
string s;
int ans;
void Query() {
	int now = 0;
	for (int i = 0; i < s.size(); i++) {
		now = t[now].id[s[i] - 'a'];
		for (int to = now; to; to = t[to].fail) {
			if (t[to].cnt == -1) break;
			ans += t[to].cnt;
			t[to].cnt = -1;
		}
	}
}