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CF 1051F The Shortest Statement

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题目描述

http://codeforces.com/problemset/problem/1051/F

简要题意：给定一个 $n$ 个点 $m$ 条边的无向连通图，边有权，有 $q$ 次询问， 每次询问给定一个起点 $s_i$ 和一个终点 $t_i$，求起点到终点的最短路

$n,m,q\le 10^5,m-n\le 20$

Solution

这张图可以转换为一棵树多连了 $20$ 条边，那么我们可以将 $s_i$ 到 $t_i$ 的最短路分成两种，第一种是只走树边，第二种是要经过一些多余的边

所以我们将这 $20$ 条边所连的 $40$ 个点，看成关键点，对于每个关键点都跑一个最短路即可

注意到关键点只有 $40$ 个，所以我们对于每个关键点跑一遍最短路即可

时间复杂度为 $O(40n\log n)$

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#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#define maxn 100010
#define maxm 100010
#define maxk 50
#define INF 1000000000000000000
#define ll long long
using namespace std;

int n, m, q; 

int fa[maxn];
void init_fa(int n) { for (int i = 1; i <= n; ++i) fa[i] = i; } 

int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }

inline void merge(int x, int y) {
    int fx = find(x), fy = find(y);
    if (fx == fy) return ;
    fa[fy] = fx;
} 

struct Edge {
    int fr, to, next, w; 
} e[maxm * 2], E[maxm]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
    e[c1].to = v; e[c1].w = w; 
    e[c1].next = head[u]; head[u] = c1++;
}

int f[maxn][21], dep[maxn]; ll dis[maxn];
void dfs(int u, int fa) {
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = e[i].w; if (v == fa) continue;
        f[v][0] = u; dep[v] = dep[u] + 1; dis[v] = dis[u] + w;
        dfs(v, u); 
    } 
}

void init_lca() {
    for (int j = 1; j <= 20; ++j)
        for (int i = 1; i <= n; ++i) f[i][j] = f[f[i][j - 1]][j - 1];
}

int get_lca(int x, int y) {
    if (dep[x] < dep[y]) swap(x, y); 
    for (int i = 20; ~i; --i)
        if (dep[f[x][i]] >= dep[y]) x = f[x][i];
    if (x == y) return x;
    for (int i = 20; ~i; --i)
        if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return f[x][0];
}

struct Queue {
    int k; ll d;

    friend bool operator < (const Queue &u, const Queue &v) { return u.d > v.d; } 
}; bool vis[maxn]; ll d[maxk][maxn];
void dijkstra(ll *dis, int s) {
    for (int i = 1; i <= n; ++i) dis[i] = INF, vis[i] = 0;
    priority_queue<Queue> Q; dis[s] = 0; Q.push({ s, dis[s] });
    while (!Q.empty()) {
        int u = Q.top().k; Q.pop();
        if (vis[u]) continue; vis[u] = 1;
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if (dis[v] > dis[u] + w) {
                dis[v] = dis[u] + w;
                Q.push({ v, dis[v] });
            }
        }
    }
}

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m; init_fa(n); set<int> S; int cnt = 0; 
    for (int i = 1; i <= m; ++i) cin >> E[i].fr >> E[i].to >> E[i].w; 
    for (int i = 1; i <= m; ++i) {
        int u = E[i].fr, v = E[i].to, w = E[i].w;
        if (find(u) == find(v)) { S.insert(u); S.insert(v); continue; }
        merge(u, v); add_edge(u, v, w); add_edge(v, u, w); 
    } dep[1] = 1; dfs(1, 0); init_lca();
    fill(head, head + maxn, -1); c1 = 0;
    for (int i = 1; i <= m; ++i) add_edge(E[i].fr, E[i].to, E[i].w), add_edge(E[i].to, E[i].fr, E[i].w);
    for (auto u : S) dijkstra(d[++cnt], u);
    cin >> q;
    while (q--) {
        int x, y; cin >> x >> y;
        ll ans = dis[x] + dis[y] - 2 * dis[get_lca(x, y)];
        for (int i = 1; i <= cnt; ++i) 
            ans = min(ans, d[i][x] + d[i][y]);
        cout << ans << "\n";
    } 
    return 0;
}