图论算法模板

//时间复杂度O((n+m)log2​n)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

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

const int maxn = 100000 + 100;
int n, m, s, dist[maxn];
struct Edge { int v, w; };
vector<Edge> g[maxn];
struct fow {
    int u, dist;
    bool operator< (const fow& f) const { return dist > f.dist; }
};
priority_queue<fow> pq;
/*
 另外一种写法：
 struct fow { int v, w; };
 struct cmp { bool operator()(fow f1, fow f2) { return f1.w > f2.w; } };
 priority_queue<fow, vector<fow>, cmp> pq;
 注意pq使用的是>来表示<
 */
inline void dijkstra(int s) {
    memset(dist, 0x3f, sizeof dist);
    pq.push(fow { s, 0 }); dist[s] = 0;
    while (!pq.empty()) {
        int u = pq.top().u, dis = pq.top().dist; pq.pop();
        if (dist[u] != dis) continue;
        //懒删除是复杂度保证，去掉这行就会TLE
        for (int i = 0; i < g[u].size(); i++) {
            int v = g[u][i].v, w = g[u][i].w;
            if (dist[u] + w < dist[v]) { dist[v] = dist[u] + w; pq.push(fow { v, dist[v] }); }
        }
    }
}

int main() {
    n = read(); m = read(); s = read();
    for (int i = 1; i <= m; i++) {
        int a, b, c; a = read(); b = read(); c = read();
        g[a].push_back(Edge { b, c });
    }
    dijkstra(s);
    for (int i = 1; i <= n; i++) cout << dist[i] << " ";
    cout << endl;
    return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

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

const int maxn = 500000 + 100, maxpow = 20;
int n, m, s, p[maxn][50], dep[maxn], _pow[50], lg[maxn];
vector<int> g[maxn * 2];

void dfs(int u, int fa) {
    for (int i = 1; (1 << i) <= dep[u]; i++) {
        p[u][i] = p[p[u][i - 1]][i - 1];
    }
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == fa) continue;
        p[v][0] = u; dep[v] = dep[u] + 1;
        dfs(v, u);
    }
}

inline int lca(int x, int y) {
    if (dep[x] < dep[y]) swap(x, y);
    while (dep[x] > dep[y]) { x = p[x][lg[dep[x] - dep[y]] - 1]; }
    if (x == y) return x;
    for (int i = lg[dep[x]] - 1; i >= 0; i--) {
        if (p[x][i] == p[y][i]) continue;
        x = p[x][i]; y = p[y][i];
    }
    return p[x][0];
}

int main() {
    ios::sync_with_stdio(false);
    n = read(); m = read(); s = read();
    for (int i = 0; i <= 31; i++) _pow[i] = 1 << i;
    for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i); //预处理log2
    for (int i = 1; i < n; i++) {
        int a = read(), b = read();
        g[a].push_back(b); g[b].push_back(a);
    }
    p[s][0] = s;
    dfs(s, -1);
    for (int i = 1; i <= m; i++) { int x = read(), y = read(); cout << lca(x, y) << endl; }
    return 0;
}