noip板子

倍增法lca

const int N = 500010;
int n, m, s;
vector<int> g[N];
void addeg(int u, int v) {
    g[u].push_back(v);
    g[v].push_back(u);
}

int d[N], anc[N][25];
void dfs(int u, int fa) {
    d[u] = d[fa] + 1;
    for (int i : g[u]) {
        if (i == fa) continue;
        anc[i][0] = u;
        dfs(i, u);
    }
}
void initlca() {
    for (int i = 1; i <= 21; ++i) {
        for (int j = 1; j <= n; ++j) {
            anc[j][i] = anc[anc[j][i-1]][i-1];
        }
    }
}
int qry(int u, int v) {
    if (d[u] < d[v]) swap(u, v);
    for (int i = 21; i >= 0; i --) {
        if (d[anc[u][i]] >= d[v]) {
            u = anc[u][i];
        }
    }
    if (v == u) return v;
    for (int i = 21; i >= 0; i --) {
        if (anc[v][i] != anc[u][i]){
            v = anc[v][i];
            u = anc[u][i];
        }
    }
    return anc[u][0];
}

Kruakal求最小生成树

const int N = 5010, M = 200010;
struct eg {
    int u, v, w;
} es[M];
bool cmp(eg a, eg b) {return a.w < b.w;}
int f[N], n, m;
int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}
int cnt, ans;
void kruskal() {
    sort(es+1, es+m+1, cmp);
    for (int i = 1; i <= m; i ++) {
        int u = find(es[i].u), v = find(es[i].v);
        if (u == v) continue;
        ans += es[i].w;
        f[u] = v;
        cnt++;
    }
}

Dijkstra

const int N = 100010;
struct nd {
    int u, dis;
    bool operator<(const nd &a) const {
        return dis > a.dis;
    }
};
struct eg {
    int v, w;
};
priority_queue<nd> q;
vector<eg> g[N];
int n, m, dis[N], s;
bool vis[N];
void addeg(int u, int v, int w) {
    g[u].push_back((eg){v, w});
}
void dij(){
    fill(dis+1, dis+n+1, INT_MAX);
    dis[s] = 0;
    q.push((nd){s, 0});
    while (!q.empty()) {
        nd now = q.top(); q.pop();
        if (vis[now.u]) continue;
        vis[now.u] = 1;
        for (eg i : g[now.u]) {
            if (dis[i.v] > (i64)dis[now.u] + i.w) {
                dis[i.v] = dis[now.u] + i.w;
                q.push((nd){i.v, dis[i.v]});
            }
        }
    }
}