c++算法竞赛常用板子集合（持续更新）
作者：互联网
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][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 pp 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
pp 数组表示基底，xx 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
ss 和 tt 为需要匹配的两个 char 类型数组。
borderiborderi 表示 tt 长度为 ii 的前缀最长的 borderborder 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;
}
}
}