模版

目录

*由于被外星人入侵，英文目录无法快速跳转

• 通用
  • 快读
  • 卡常
• 数论
  • 快速幂
  • 线性筛素数
  • 最大公约数
  • 拓展欧几里得
  • 费马小定理
  • 欧拉函数
• 数据结构
  • 并查集
  • 线段树1
  • 线段树2
  • 线段树3
  • 树状数组1
  • 树状数组2
• 图论
  • Floyd
  • 传递闭包
  • Dijkstra1
  • Dijkstra2
  • Bellman-Ford
  • SPFA
  • 倍增求LCA
• 镇楼图

通用

快读

仅适用于整数的快读：

inline long long read(){
    long long s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * w;
}

__int128

inline __int128 read(){
    __int128 s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * w;
}

inline void print(__int128 x){
	if (x == 0)
		return;
	if (x < 0)
		putchar('-'), x = -x;
	print(x / 10);
	putchar(x % 10 + '0');
}

void work(){
	__int128 a;
	a = read();
	if (a == 0)
		puts("0");
	else
		print(a);
}

卡常

dfs必备

#pragma GCC target("sse,sse2,sse3,sse4.1,sse4.2,popcnt,abm,mmx,avx")
#pragma comment(linker,"/STACK:102400000,102400000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")

数论

快速幂

计算:\(a ^ b \% p\)

int power(int a, int b, int p){
    int ans = 1 % p;
    for (; b; b >>= 1){
        if (b & 1) ans = (long long)ans * a % p;
        a = (long long)a * a % p;
    }
    return ans;
}

线性筛素数

int n, cnt, p[100000010];
bool v[100000010];
void prime(){
    v[1] = 1;
    for (int i = 2; i <= n; i++){
        if (!v[i]) p[++cnt] = i;
        for (int j = 1; p[j] * i <= n; j++){
            v[p[j] * i] = 1;
            if (i % p[j] == 0) break;
        }
    }
}

最大公约数

方法:辗转相除

int gcd (int a, int b){
    if (b == 0) return a;
    return gcd (b, a % b);
    //return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b){
    return a / gcd(a, b) * b;
}

拓展欧几里得

int exgcd(int a, int b, int &x, int &y) {
    if (b == 0){
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

void work1(){
    int a, b, x, y, d;
    scanf("%d%d", &a, &b);
    d = exgcd(a, b, x, y);
    printf("gcd(a, b) = %d\n", d);
    printf("x = %d, y = %d\n", x, y);
}

void work2(){ //乘法逆元
    int n = read(), p = read();
    for (int i = 1; i <= n; i++){
        int x, y;
        exgcd(i, p, x, y);
        x = (x % p + p) % p;
        printf("%d\n", x); //x是a在mod p下的逆元
    }
}

费马小定理

公式: \(a^{p-1} \equiv 1 \pmod{p} (a \nmid p)\) 其中 \(p\) 是素数

应用: 乘法逆元 \(a \times a^{p-2} \equiv 1 \pmod{p} (a \nmid p)\) 其中 \(p\) 是素数

时间复杂度: \(\Theta(\log n)\)

欧拉函数

int n, cnt;
int phi[1000010], p[1000010];
bool vis[1000010];
void getphi(){
    phi[1] = 1;
    for (int i = 2; i <= n; i++){
        if (!vis[i]) p[++cnt] = i, phi[i] = i-1;
        for (int j = 1; j <= cnt; j++){
            if (i * p[j] > n) break;
            vis[i * p[j]] = 1;
            if (i % p[j] == 0){
                phi[i*p[j]] = phi[i] * p[j];
                break;
            }
            phi[i*p[j]] = phi[i] * (p[j]-1);
        }
    }
}

数据结构

并查集

const int MAXN = 100010;
int fa[MAXN], n; //有n个点
//int size[MAXN]
int find(int x){
    if (x == fa[x]) return x;
    return fa[x] = find(fa[x]);
    //return (x == fa[x]) ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y){
    x = find(x), y = find(y);
    fa[x] = y;
    //size[y] += size[x];
}
void init(){
    for (int i = 1; i <= n; i++) fa[i] = i;
}

线段树1

实现一个数据结构支持区间加，区间求和，具体要求如下

1. 1 l r k 把区间 \([l, r]\) 中的每一项增加 \(k\)
2. 2 l r 求区间 \([l, r]\) 中的每一个数的和

const int MAXN = 100010;
int a[MAXN], n, m;
struct SegmentTree{
    int l, r;
    long long sum, add;
    #define l(x) t[x].l
    #define r(x) t[x].r
    #define sum(x) t[x].sum
    #define add(x) t[x].add
}t[MAXN*4];

void build(int p, int l, int r){
    l(p) = l, r(p) = r;
    if (l == r){
        sum(p) = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p*2, l, mid);
    build(p*2+1, mid+1, r);
    sum(p) = sum(p*2) + sum(p*2+1);
}

void spread(int p){
    if (add(p)){
        sum(p*2) += add(p) * (r(p*2) - l(p*2) + 1);
        sum(p*2+1) += add(p) * (r(p*2+1) - l(p*2+1) + 1);
        add(p*2) += add(p);
        add(p*2+1) += add(p);
        add(p) = 0;
    }
}

void change(int p, int l, int r, int d){
    if(l <= l(p) && r >= r(p)){
        sum(p) += (long long)d * (r(p) - l(p) + 1);
        add(p) += d;
        return;
    }
    spread(p);
    int mid = (l(p) + r(p)) / 2;
    if (l <= mid) change(p*2, l, r, d);
    if (r > mid) change(p*2+1, l, r, d);
    sum(p) = sum(p*2) + sum(p*2+1);
}

long long ask(int p, int l, int r){
    if (l <= l(p) && r >= r(p)) return sum(p);
    spread(p);
    int mid = (l(p)+r(p)) >> 1;
    long long ans = 0;
    if (l <= mid) ans += ask(p*2, l, r);
    if (r > mid) ans += ask(p*2+1, l ,r);
    return ans;
}

void work(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    build(1, 1, n);
    while (m--){
        int flag = read();
        if (flag == 1){
            int x = read(), y = read(), k = read();
            change(1, x, y, k);
        }
        else{
            int x = read(), y = read();
            printf("%lld\n", ask(1, x, y));
        }
    }
}

线段树2

实现一个数据结构支持区间加，区间求和，具体要求如下

1. 1 l r k 把区间 \([l, r]\) 中的每一项增乘 \(k\)
2. 2 l r k 把区间 \([l, r]\) 中的每一项增加 \(k\)
3. 3 l r 求区间 \([l, r]\) 中的每一个数的和

const int MAXN = 100010;
int a[MAXN], n, m, mo;
struct SegmentTree{
    int l, r;
    long long sum, add = 0, mul = 1;
    #define l(x) t[x].l
    #define r(x) t[x].r
    #define sum(x) t[x].sum
    #define add(x) t[x].add
    #define mul(x) t[x].mul
}t[MAXN*4];

void build(int p, int l, int r){
    l(p) = l, r(p) = r;
    if (l == r){
        sum(p) = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p*2, l, mid);
    build(p*2+1, mid+1, r);
    sum(p) = sum(p*2) + sum(p*2+1);
}

void spread(int p){
    //更新每段的和
    sum(p*2) = (sum(p*2) * mul(p) + add(p) * (r(p*2) - l(p*2) + 1)) % mo;
    sum(p*2+1) = (sum(p*2+1) * mul(p) + add(p) * (r(p*2+1) - l(p*2+1) + 1)) % mo;
    //先更新乘法懒标记，后更新加法懒标记
    mul(p*2) = (mul(p*2) * mul(p)) % mo;
    mul(p*2+1) = (mul(p*2+1) * mul(p)) % mo;
    add(p*2) = (add(p*2) * mul(p) + add(p)) % mo;
    add(p*2+1) = (add(p*2+1) * mul(p) + add(p)) % mo;
    //懒标记归位
    mul(p) = 1;
    add(p) = 0;
}

void change_add(int p, int l, int r, int d){
    if(l <= l(p) && r >= r(p)){
        sum(p) += d * (r(p) - l(p) + 1), sum(p) %= mo;
        add(p) += d, add(p) %= mo;
        return;
    }
    spread(p);
    int mid = (l(p) + r(p)) / 2;
    if (l <= mid) change_add(p*2, l, r, d);
    if (r > mid) change_add(p*2+1, l, r, d);
    sum(p) = sum(p*2) + sum(p*2+1), sum(p) %= mo;
}

void change_mul(int p, int l, int r, int d){
    if(l <= l(p) && r >= r(p)){
        sum(p) *= d, sum(p) %= mo;
        mul(p) *= d, mul(p) %= mo;
        add(p) *= d, add(p) %= mo;
        return;
    }
    spread(p);
    int mid = (l(p) + r(p)) / 2;
    if (l <= mid) change_mul(p*2, l, r, d);
    if (r > mid) change_mul(p*2+1, l, r, d);
    sum(p) = sum(p*2) + sum(p*2+1), sum(p) %= mo;
}

long long ask(int p, int l, int r){
    if (l <= l(p) && r >= r(p)) return sum(p) % mo;
    spread(p);
    int mid = (l(p)+r(p)) >> 1;
    long long ans = 0;
    if (l <= mid) ans += ask(p*2, l, r), ans %= mo;
    if (r > mid) ans += ask(p*2+1, l ,r), ans %= mo;
    return ans % mo;
}

void work(){
    n = read(), m = read(), mo = read();
    for (int i = 1; i <= n; i++)
        a[i] = read() % mo;
    build(1, 1, n);
    while (m--){
        int flag = read();
        if (flag == 1){
            int x = read(), y = read(), k = read() % mo;
            change_mul(1, x, y, k);
        }
        else if (flag == 2){
            int x = read(), y = read(), k = read() % mo;
            change_add(1, x, y, k);
        }
        else{
            int x = read(), y = read();
            printf("%lld\n", ask(1, x, y) % mo);
        }
    }
}

线段树3

实现一个数据结构支持单点修改，区间求最大子段和，具体要求如下

1. 1 l r 求区间 \([l, r]\) 中的最大子段和 \(k\)
2. 2 x k 把第 \(x\) 项改为 \(k\)

const int MAXN = 500010, inf = 0x7fffffff;
int a[MAXN], n, m;
struct SegmentTree{
    int l, r, sum;
    int lmax, rmax, mmax;
    #define l(x) t[x].l
    #define r(x) t[x].r
    #define sum(x) t[x].sum
    #define m(x) t[x].mmax
    #define lm(x) t[x].lmax
    #define rm(x) t[x].rmax
}t[MAXN*4];

void update(int p){
    sum(p) = sum(p<<1) + sum(p<<1|1);
    lm(p) = max(sum(p<<1)+lm(p<<1|1), lm(p<<1));
    rm(p) = max(sum(p<<1|1)+rm(p<<1), rm(p<<1|1));
    m(p) = max(max(m(p<<1), m(p<<1|1)), rm(p<<1)+lm(p<<1|1));
}

void build(int p, int l, int r){
    l(p) = l, r(p) = r;
    if (l == r){
        sum(p) = m(p) = lm(p) = rm(p) = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p<<1, l, mid);
    build(p<<1|1, mid+1, r);
    update(p);
}

void change(int p, int x, int d){
    if(l(p)== r(p)){
        sum(p) = m(p) = lm(p) = rm(p) = d;
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if (x <= mid) change(p<<1, x, d);
    if (x > mid) change(p<<1|1, x, d);
    update(p);
}

SegmentTree ask(int p, int l, int r){
    if (l <= l(p) && r(p) <= r) return t[p];
    int mid = (l(p) + r(p)) >> 1;
    if(r <= mid) return ask(p<<1, l, r);
    if(l > mid) return ask(p<<1|1, l, r);
    SegmentTree x = ask(p<<1, l, r), y = ask(p<<1|1, l, r), re;
    re.sum = x.sum + y.sum;
    re.lmax = max(x.sum+y.lmax, x.lmax);
    re.rmax = max(y.sum+x.rmax, y.rmax);
    re.mmax = max(max(x.mmax, y.mmax), x.rmax+y.lmax);
    return re;
}

void work(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    build(1, 1, n);
    while (m--){
        int flag = read();
        if (flag == 1){
            int x = read(), y = read();
            if (x > y) swap(x, y);
            printf("%d\n", ask(1, x, y).mmax);
        }
        else{
            int x = read(), y = read();
            change(1, x, y);
        }
    }
}

树状数组1

实现一个数据结构支持单点加，区间求和，具体要求如下

1. 1 i k 把第 \(i\) 项增加 \(k\)
2. 2 l r 求区间 \([l, r]\) 中的每一个数的和

const int MAXN = 600060;
int n, m, a[MAXN], c[MAXN];

void add(int x, int y){
    for (; x <= n; x += x & -x) c[x] += y;
}

int ask(int x){
    int ans = 0;
    for (; x; x -= x & -x)
        ans += c[x];
    return ans;
}

void work(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    for (int i = 1; i <= n; i++)
        add(i, a[i]);
    while (m--){
        int flag = read();
        if (flag == 1){
            int x = read(), k = read();
            add(x, k);
        }
        else{
            int x = read(), y = read();
            printf("%d\n", ask(y) - ask(x-1));
        }
    }
}

树状数组2

实现一个数据结构支持区间加，区间求和，具体要求如下

1. 1 l r k 把区间 \([l, r]\) 中的每一项增加 \(k\)
2. 2 l r 求区间 \([l, r]\) 中的每一个数的和

const int MAXN = 600010;
int a[MAXN], n, m;
long long c[2][MAXN], sum[MAXN];

long long ask(int k, int x){
    long long ans = 0;
    for (; x; x -= x & -x) ans += c[k][x];
    return ans;
}

void add(int k, int x, int y){
    for (; x <= n; x += x & -x) c[k][x] += y;
}

void work(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i] = read(), sum[i] = sum[i-1] + a[i];
    while (m--){
        int falg = read();
        if (falg == 1){
            int x = read(), y = read(), k = read();
            add(0, x, k);
            add(0, y+1, -k);
            add(1, x, x * k);
            add(1, y+1, -(y+1) * k);
        }
        else{
            int x = read(), y = read();
            long long ans = sum[y] + (y+1) * ask(0, y) - ask(1, y);
            ans -= sum[x-1] + x * ask(0, x-1) - ask(1, x-1);
            printf("%lld\n", ans);
        }
    }
}

图论

Floyd

基本思想: 动态规划

时间复杂度: \(\Theta(n^3)\)

存图方式: 邻接矩阵

Code

int n, m, d[3333][3333];

void Floyd(){
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (i != j && i != k && j != k)
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

void work(){
    n = read(), m = read();
    memset(d, 0x3f, sizeof(d));
    for (int i = 1; i <= n; i++)
        d[i][i] = 0;
    for (int i = 1; i <= m; i++){
        int x = read(), y = read(), z = read();
        d[x][y] = min(d[x][y], z);
    }
    Floyd();
}

传递闭包

基本思想: Floyd

时间复杂度: \(\Theta(n^3)\)

存图方式: 邻接矩阵

Code

int n, m, d[3333][3333];

void Floyd(){
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (i != j && i != k && j != k)
                    d[i][j] |= d[i][k] & d[k][j];
}

void work(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        d[i][i] = 1;
    for (int i = 1; i <= m; i++){
        int x = read(), y = read();
        d[x][y] = d[y][x] = 1;
    }
    Floyd();
}

Dijkstra1

基本思想: 贪心

时间复杂度: \(\Theta(n^2)\)

存图方式: 邻接矩阵

Code

const int MAXN = 3030;
int a[MAXN][MAXN], d[MAXN], n, m, s;
bool v[MAXN];

void Dijkstra(){
    memset(d, 0x3f, sizeof(d));
    memset(v, false, sizeof(v));
    d[s] = 0;
    for (int i = 1; i < n; i++){
        int x = 0;
        for (int j = 1; j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x]))
                x = j;
        v[x] = true;
        for (int y = 1; y <= n; y++)
            d[y] = min(d[y], d[x] + a[x][y]);
    }
}

void work(){
    memset(a, 0x3f, sizeof(a));
    n = read(), m = read(), s = read();
    for (int i = 1; i <= n; i++)
        a[i][i] = 0;
    for (int i = 1; i <= m; i++){
        int x = read(), y = read(), z = read();
        a[x][y] = min(a[x][y], z);
    }
    Dijkstra();
}

Dijkstra2

基本思想: 贪心

时间复杂度: \(\Theta(m \log n)\)

存图方式: 链表

Code

const int MAXN = 1e5 + 10, MAXM = 1e6 +10;
int head[MAXN], ver[MAXM], edge[MAXM], Next[MAXM], d[MAXN];
bool v[MAXN];
int n, m, s, tot;
priority_queue <pair<int, int> > q;

void add(int u, int v, int w){
    tot++;
    ver[tot] = v, edge[tot] = w;
    Next[tot] = head[u], head[u] = tot;
}

void Dijkstra(int st){
    memset(d, 0x3f, sizeof(d));
    memset(v, 0, sizeof(v));
    d[st] = 0;
    q.push(make_pair(0, st));
    while (!q.empty()){
        int x = q.top().second;
        q.pop();
        if (v[x]) continue;
        v[x] = 1;
        for (int i = head[x]; i; i = Next[i]){
            int y = ver[i], z = edge[i];
            if (d[y] > d[x] + z){
                d[y] = d[x] + z;
                q.push (make_pair(-d[y], y));
            }
        }
    }
}

void work(){
    n = read(), m = read(), s = read();
    for (int i = 1; i <= m; i++){
        int u = read(), v = read(), w = read();
        add(u, v, w);
    }
    Dijkstra(s);
}

Bellman-Ford

即是没有队列优化的SPFA, 中文名: 福特算法, 不常用

基本思想: 迭代

时间复杂度: \(\Theta(n m)\)

存图方式: 数组

Code

const int MAXN = 100010;
int n, m;
int d[MAXN], w[MAXN], ax[MAXN], ay[MAXN];
bool v[MAXN];

void Ford(int s){
    d[s] = 0;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            d[ax[j]] = min(d[ax[j]], d[ay[j]] + w[j]);
            d[ay[j]] = min(d[ay[j]], d[ax[j]] + w[j]);
        }
    }
}

void work(){
    memset(d, 0x3f, sizeof(d));
    memset(w, 0x3f, sizeof(w));
    n = read(), m = read();
    int s = read();
    for (int i = 1; i <= m; i++){
        ax[i] = read(), ay[i] = read();
        w[i] = read();
    }
    Ford(s);
}

SPFA

关于SPFA, 它死了……

基本思想: Ford(即队列优化的Ford算法)

时间复杂度: \(\Theta(k m)\), 通常情况下 \(k = 2\), 最坏情况下 \(k = n\)

Code

const int MAXN = 150000;
int head[MAXN], ver[MAXN], edge[MAXN], Next[MAXN], d[MAXN];
int n, m, tot, s;
bool v[MAXN];
queue <int> q;

void add(int x, int y, int z){
    tot++;
    ver[tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;
}

void spfa(int i){
    memset(d, 0x3f, sizeof(d));
    //memset(v, 0, sizeof(v));
    d[i] = 0;
    v[i] = 1;
    q.push(i);
    while(!q.empty()){
        int x = q.front();
        q.pop();
        v[x] = 0;
        for (int i = head[x]; i; i = Next[i]){
            int y = ver[i];
            long long z = edge[i];
            if (d[y] > d[x]+z){
                d[y] = d[x]+z;
                if (!v[y]) q.push(y), v[y] = 1;
            }
        }
    }
}

void work(){
    n = read(), m = read(), s = read();
    for (int i = 1; i <= m; i++){
        int x = read(), y = read(), z = read();
        add(x, y, z), add(y, x, z);
    }
    spfa(s);
}

倍增求LCA

const int SIZE = 500010;
int n, m, s, tot, t;
int f[SIZE][41], d[SIZE];
int ver[2*SIZE], Next[2*SIZE], head[SIZE];
queue <int> q;

void add(int x, int y){
    tot++;
    ver[tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
}

void bfs(){
    q.push(s);
    d[s] = 1;
    while (q.size()){
        int x = q.front();
        q.pop();
        for (int i = head[x]; i; i = Next[i]){
            int y = ver[i];
            if (d[y]) continue;
            d[y] = d[x] + 1;
            f[y][0] = x;
            for (int j = 1; j <= t; j++)
                f[y][j] = f[f[y][j-1]][j-1];
            q.push(y);
        }
    }
}

int lca(int x, int y){
    if (d[x] > d[y]) swap(x, y);
    for (int i = t; i >= 0; i--)
        if (d[f[y][i]] >= d[x])
            y = f[y][i];
    if (x == y) return x;
    for (int i = t; i >= 0; i--)
        if (f[x][i] != f[y][i])
            x = f[x][i], y = f[y][i];
    return f[x][0];
}

void work(){
    n = read(), m = read(), s = read();
    t = (int)(log(n) / log(2)) + 1;
    for (int i = 1; i <= n-1; i++){
        int x = read(), y = read();
        add(x, y), add(y, x);
    }
    bfs();
    for (int i = 1; i <= m; i++){
        int x = read(), y = read();
        printf("%d\n", lca(x, y));
    }
}

镇楼图