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
快速读入
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#define ll long long
inline ll read() {
ll s = 0, w = 1;
char c = getchar();
while (c < '0' || c > '9') {if (c == '-') w = -1; c = getchar();}
while (c >= '0' && c <= '9') s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
return s * w;
}
高精加法
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struct node {
int c[N], l;
node() {l = 0;}
void print() {
for (int i = l; i >= 1; i--) printf("%d", c[i]);
printf("\n");
}
} a, b, ans;
node operator +(const node &a, const node &b) {
node res; res.l = max(a.l, b.l);
for (int i = 1; i <= res.l; i++) res.c[i] = a.c[i] + b.c[i];
for (int i = 1; i < res.l; i++) res.c[i + 1] += res.c[i] / 10, res.c[i] %= 10;
while (res.c[res.l] > 9) res.c[res.l + 1] = res.c[res.l] / 10, res.c[res.l] %= 10, res.l++;
while (res.c[res.l] == 0 && res.l > 1) res.l--;
return res;
}
高精减法
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struct node {
int c[N], l;
node() {l = 0;}
void print() {
for (int i = l; i >= 1; i--) printf("%d", c[i]);
printf("\n");
}
} a, b, ans;
node operator -(const node &a, const node &b) {
node res; res.l = max(a.l, b.l);
for (int i = 1; i <= res.l; i++) res.c[i] = a.c[i] - b.c[i];
for (int i = 1; i < res.l; i++)
if (res.c[i] < 0) res.c[i] += 10, res.c[i + 1]--;
while (res.c[res.l] == 0 && res.l > 1) res.l--;
return res;
}
bool operator >(const node &a, const node &b) {
if (a.l != b.l) return a.l > b.l;
for (int i = a.l; i >= 1; i--)
if (a.c[i] != b.c[i]) return a.c[i] > b.c[i];
return 0;
}
if (b > a) printf("-"), ans = b - a;
else ans = a - b;
高精乘法
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struct node {
int c[N], l;
node() {l = 0;}
void print() {
for (int i = l; i >= 1; i--) printf("%d", c[i]);
printf("\n");
}
} a, b, ans;
node operator *(const node &a, const node &b) {
node res; res.l = a.l + b.l - 1;
for (int i = 1; i <= a.l; i++)
for (int j = 1; j <= b.l; j++)
res.c[i + j - 1] += a.c[i] * b.c[j];
for (int i = 1; i < res.l; i++) res.c[i + 1] += res.c[i] / 10, res.c[i] %= 10;
while (res.c[res.l] > 9) res.c[res.l + 1] += res.c[res.l] / 10, res.c[res.l] %= 10, res.l++;
while (res.c[res.l] == 0 && res.l > 1) res.l--;
return res;
}
树状数组
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// 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
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// 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;
动态开点线段树
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// 此处为区间修改区间查询区间和的动态开点线段树。
struct SegmentTree {
int l[N << 1], r[N << 1], ls[N << 1], rs[N << 1], cnt; ll sum[N << 1], lazy[N << 1];
void update(int rt) {sum[rt] = sum[ls[rt]] + sum[rs[rt]];}
int build(int L, int R) {
cnt++;
l[cnt] = L, r[cnt] = R, sum[cnt] = 0, lazy[cnt] = 0;
return cnt;
}
void pushdown(int rt) {
if (lazy[rt] == 0) return ;
lazy[ls[rt]] += lazy[rt], lazy[rs[rt]] += lazy[rt];
sum[ls[rt]] += lazy[rt] * (r[ls[rt]] - l[ls[rt]] + 1);
sum[rs[rt]] += lazy[rt] * (r[rs[rt]] - l[rs[rt]] + 1);
lazy[rt] = 0;
}
void change(int rt, int L, int R, int x) {
if (r[rt] < L || R < l[rt]) return ;
if (L <= l[rt] && r[rt] <= R) {sum[rt] += x * (r[rt] - l[rt] + 1); lazy[rt] += x; return ;}
pushdown(rt);
int mid = l[rt] + r[rt] >> 1;
if (L <= mid) {
if (!ls[rt]) ls[rt] = build(l[rt], mid);
change(ls[rt], L, R, x);
}
if (mid + 1 <= R) {
if (!rs[rt]) rs[rt] = build(mid + 1, r[rt]);
change(rs[rt], L, R, x);
}
update(rt);
}
ll query(int rt, int L, int R) {
if (r[rt] < L || R < l[rt]) return 0;
if (L <= l[rt] && r[rt] <= R) return sum[rt];
pushdown(rt);
int mid = l[rt] + r[rt] >> 1; ll res = 0;
if (L <= mid) {
if (!ls[rt]) ls[rt] = build(l[rt], mid);
res += query(ls[rt], L, R);
}
if (mid + 1 <= R) {
if (!rs[rt]) rs[rt] = build(mid + 1, r[rt]);
res += query(rs[rt], L, R);
}
return res;
}
} tree;
扫描线
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//此处为统计矩阵总覆盖面积的扫描线
const int N = 100010;
int n;
struct node {
int l, r, h, tag;
bool operator <(const node &a) {return h < a.h;}
} lne[N << 1]; int lcnt;
int tx[N << 1], tcnt;
struct SegmentTree {
int l[N << 4], r[N << 4]; ll sum[N << 4], len[N << 4];
void update(int rt) {
if (sum[rt]) len[rt] = tx[r[rt] + 1] - tx[l[rt]];
else len[rt] = len[rt << 1] + len[rt << 1 | 1];
}
void build(int rt, int L, int R) {
l[rt] = L, r[rt] = R;
if (L == R) {sum[rt] = len[rt] = 0; return ;}
int mid = L + R >> 1;
build(rt << 1, L, mid), build(rt << 1 | 1, mid + 1, R);
update(rt);
}
void add(int rt, int L, int R, int x) {
if (tx[r[rt] + 1] <= L || R <= tx[l[rt]]) return ;
if (L <= tx[l[rt]] && tx[r[rt] + 1] <= R) {sum[rt] += x; update(rt); return ;}
add(rt << 1, L, R, x), add(rt << 1 | 1, L, R, x);
update(rt);
}
} tree;
n = read();
for (int i = 1; i <= n; i++) {
int xa = read(), ya = read(), xb = read(), yb = read();
tx[++tcnt] = xa, tx[++tcnt] = xb;
lne[++lcnt].l = xa, lne[lcnt].r = xb, lne[lcnt].h = ya, lne[lcnt].tag = 1;
lne[++lcnt].l = xa, lne[lcnt].r = xb, lne[lcnt].h = yb, lne[lcnt].tag = -1;
}
sort(tx + 1, tx + 1 + tcnt); tcnt = unique(tx + 1, tx + 1 + tcnt) - tx - 1;
sort(lne + 1, lne + 1 + lcnt);
tree.build(1, 1, tcnt - 1);
for (int i = 1; i < lcnt; i++) {
tree.add(1, lne[i].l, lne[i].r, lne[i].tag);
ans += tree.len[1] * (lne[i + 1].h - lne[i].h);
}
printf("%lld\n", ans);
堆
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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
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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
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// 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
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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 (Tarjan法) 更多方法
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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 (堆优化Dijkstra)
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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
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int dfn[N], low[N], dcnt, p[N], sum[N];
bool instack[N];
stack <int> s;
void dfs(int x) {
dfn[x] = low[x] = ++dcnt;
instack[x] = 1; s.push(x);
for (int i = oldg.last[x]; i; i = oldg.e[i].next) {
int v = oldg.e[i].to;
if (dfn[v]) {
if (instack[v]) low[x] = min(low[x], dfn[v]);
continue;
}
dfs(v);
low[x] = min(low[x], low[v]);
}
if (low[x] >= dfn[x]) {
p[x] = x, sum[x] = a[x];
instack[x] = 0;
while (s.top() != x) {
int v = s.top(); s.pop();
instack[v] = 0;
p[v] = x, sum[x] += a[v];
}
s.pop();
}
}
for (int i = 1; i <= m; i++) u[i] = read(), v[i] = read();
for (int i = 1; i <= m; i++) oldg.addedge(u[i], v[i]);
for (int i = 1; i <= n; i++)
if (!dfn[i]) dfs(i);
for (int i = 1; i <= m; i++)
if (p[u[i]] != p[v[i]]) newg.addedge(p[u[i]], p[v[i]]);
点双连通分量
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int dfn[N], low[N], tcnt;
vector <int> ans[N]; int anscnt;
stack <int> s;
void dfs(int x, int fa) {
dfn[x] = low[x] = ++tcnt;
if (x == fa && !last[x]) {
ans[++anscnt].emplace_back(x);
return ;
}
s.push(x);
for (int i = last[x]; i; i = e[i].next) {
int v = e[i].to;
if (dfn[v]) {low[x] = min(low[x], dfn[v]); continue;}
dfs(v, x);
low[x] = min(low[x], low[v]);
if (low[v] >= dfn[x]) {
anscnt++;
int now;
do {
now = s.top(); s.pop();
ans[anscnt].emplace_back(now);
} while (now != v);
ans[anscnt].emplace_back(x);
}
}
}
for (int i = 1; i <= m; i++) {
int x = read(), y = read();
if (x == y) continue;
addedge(x, y), addedge(y, x);
}
for (int i = 1; i <= n; i++)
if (!dfn[i]) dfs(i, i);
printf("%d\n", anscnt);
for (int i = 1; i <= anscnt; i++) {
printf("%d ", ans[i].size());
for (int x: ans[i]) printf("%d ", x);
puts("");
}
边双连通分量
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int low[N << 1], dfn[N << 1], tcnt;
vector <int> ans[N]; int anscnt;
stack <int> s;
void dfs(int x, int edg) {
dfn[x] = low[x] = ++tcnt;
s.push(x);
for (int i = last[x]; i; i = e[i].next) {
int v = e[i].to;
if (dfn[v]) {if (i != (edg ^ 1)) low[x] = min(low[x], dfn[v]); continue;}
dfs(v, i);
low[x] = min(low[x], low[v]);
if (low[v] > dfn[x]) {
anscnt++;
int now;
do {
now = s.top(); s.pop();
ans[anscnt].emplace_back(now);
} while (now != v);
}
}
}
for (int i = 1; i <= n; i++)
if (!dfn[i]) {
addedge(i + n, i), addedge(i, i + n);
dfs(i + n, 0);
}
printf("%d\n", anscnt);
for (int i = 1; i <= anscnt; i++) {
printf("%d ", ans[i].size());
for (int x: ans[i]) printf("%d ", x);
puts("");
}
欧拉路径
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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;
}
网络最大流（dinic）
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int dep[N], cur[N];
queue <int> q;
bool bfs() {
while (q.size()) q.pop();
for (int i = 1; i <= n; i++) dep[i] = -1; dep[s] = 0;
cur[s] = last[s]; q.push(s);
while (q.size()) {
int x = q.front(); q.pop();
for (int i = last[x]; i; i = e[i].next) {
int v = e[i].to;
if (dep[v] != -1) continue;
if (e[i].w == 0) continue;
dep[v] = dep[x] + 1;
cur[v] = last[v];
if (v == t) return 1;
q.push(v);
}
}
return 0;
}
ll dfs(int x, ll tp) {
if (x == t) return tp;
ll res = 0;
for (int i = cur[x]; i && res < tp; i = e[i].next) {
cur[x] = i;
int v = e[i].to;
if (dep[v] != dep[x] + 1) continue;
if (e[i].w == 0) continue;
ll now = dfs(v, min(e[i].w, tp - res));
if (now == 0) dep[v] = -1;
e[i].w -= now, e[i ^ 1].w += now, res += now;
}
return res;
}
ll dinic() {
ll res = 0, now;
while (bfs())
while (now = dfs(s, INF)) res += now;
return res;
}
乘法逆元
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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
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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
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struct Martix {
int n, m;
ll a[N][N];
void clear() {memset(a, 0, sizeof(a));}
void init() {clear(); for (int i = 1; i <= n; i++) a[i][i] = 1;}
Martix operator *(const Martix b) const {
Martix res; res.clear(); res.n = n, res.m = b.m;
for (int i = 1; i <= res.n; i++)
for (int j = 1; j <= res.m; j++)
for (int k = 1; k <= m; k++)
res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j] % mod) % mod;
return res;
}
Martix operator ^(ll x) const {
Martix res, a = *this; res.n = res.m = n, res.init();
while (x) {
if (x & 1) res = res * a;
a = a * a;
x >>= 1;
}
return res;
}
} a;
线性基
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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
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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
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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
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// 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。
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
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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;
}
}
}
字符串哈希
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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
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// 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。
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++;
}
