cf1051F. The Shortest Statement(最短路)

题意

题目链接

题意：给出一张无向图，每次询问两点之间的最短路，满足$m - n <= 20$

$n, m, q \leqslant 10^5$

sol

非常好的一道题。

首先建出一个dfs树。

因为边数-点数非常少，所以我们可以对于某些非树边特殊考虑。

具体做法是：对于非树边连接的两个点，暴力求出它们到所有点的最短路

对于询问的$(x, y)$

用树上的边，以及非树边连接的点到他们的最短路之和更新答案

由于边数的限制，非树边连接的点不会超过$2*(m - (n - 1)) = 42$个

#include<bits/stdc++.h>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long 
using namespace std;
const int maxn = 2 * 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, q, head[maxn], num = 0, vis[maxn], fa[maxn][21], dep[maxn], happen[maxn];
ll dis[50][maxn], tdis[maxn];
vector<int> p;
struct edge {
    ll u, v, w, f, nxt;
}e[maxn];
inline void addedge(int x, int y, int z) {
    e[num] = (edge) {x, y, z, 0, head[x]};
    head[x] = num++;
}
void dfs(int x, int _fa) {
    vis[x] = 1; dep[x] = dep[_fa] + 1;
    for(int i = head[x]; ~i; i = e[i].nxt) {
        int to = e[i].v;
        if(vis[to]) continue;
        e[i].f = e[i ^ 1].f = 1;
        tdis[to] = tdis[x] + (ll)e[i].w;
        fa[to][0] = x;
        dfs(to, x);
    }
} 
void dij(int x, int id) {
    memset(dis[id], 0x7f, sizeof(dis[id])); dis[id][x] = 0;
    memset(vis, 0, sizeof(vis));
    priority_queue<pair> q; q.push(mp(0, x));
    while(!q.empty()) {
        int p = q.top().se; q.pop();
        if(vis[p]) continue;
        for(int i = head[p]; ~i; i = e[i].nxt) {
            int to = e[i].v;
            if(dis[id][to] > dis[id][p] + e[i].w && (!vis[to])) 
                dis[id][to] = dis[id][p] + e[i].w, q.push(mp(-dis[id][to], to));
        }
    }
}
void pre() {
    for(int j = 1; j <= 20; j++)
        for(int i = 1; i <= n; i++)
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
}
int lca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 20; i >= 0; i--)
        if(dep[fa[x][i]] >= dep[y]) x = fa[x][i];
    if(x == y) return x;
    for(int i = 20; i >= 0; i--)
        if(fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}
main() {
//    freopen("a.in", "r", stdin);
    memset(head, -1, sizeof(head));
    n = read(); m = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), z = read();
        addedge(x, y, z);
        addedge(y, x, z);
    }
    dfs(1, 0);
    for(int i = 0; i < num; i++) 
        if(!e[i].f) {
            if(!happen[e[i].u]) p.push_back(e[i].u), happen[e[i].u] = 1;
            if(!happen[e[i].v]) p.push_back(e[i].v), happen[e[i].v] = 1;
        }
            
    for(int i = 0; i < p.size(); i++) 
        dij(p[i], i);
    pre();
    int q = read();
    while(q--) {
        int x = read(), y = read();
        ll ans = tdis[x] + tdis[y] - 2 * tdis[lca(x, y)];
        for(int i = 0; i < p.size(); i++) 
            ans = min(ans, dis[i][x] + dis[i][y]);
        cout << ans << endl;
    }
    return 0;
}