题目链接

题意：

给 n 个点，n 条边（3? <= n <= 100000），q 次询问，每次求 u 到 v 之间的最短路。没有重边和自环。

题解：

显然 O(n2logn) 是会超时的。由于只给了n条边，那么去掉一条边就是一棵树了。这时可以用LCA求解。我看这位大佬的代码学的。

用并查集即可将成环的一条边从图中除去。之后的每次查询只需要判断经过这条边和不经过这条边哪个更小即可。具体看代码。

代码：

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <bitset>
using namespace std;

typedef long long ll;
#define int ll
#define INF 0x3f3f3f3f3f3f3f3f
#define MAXM 32 + 10
#define MAXN 300000 + 10
const ll mod = 1e9 + 7;
#define P pair<int, int>
#define fir first
#define sec second

int t, n, q;
vector<P> G[MAXN];
int father[MAXN];
int dis[MAXN], dep[MAXN];
int gfa[MAXM][MAXN];

int get_root(int x)
{
    if(x == father[x]) return x;
    return father[x] = get_root(father[x]);
}

void dfs(int now, int fa, int d)
{
    gfa[0][now] = fa;
    dep[now] = d;
    for(int i = 0; i < G[now].size(); i ++) {
        if(G[now][i].sec != fa) {
            dis[G[now][i].sec] = dis[now] + G[now][i].fir;
            dfs(G[now][i].sec, now, d + 1);
        }
    }
}

void init()
{
    dfs(1, -1, 0);
    for(int k = 0; k + 1 < 32; k ++) {
        for(int v = 1; v <= n; v ++) {
            if(gfa[k][v] == -1) gfa[k+1][v] = -1;
            else gfa[k+1][v] = gfa[k][gfa[k][v]];
        }
    }
}

int LCA(int u, int v)
{
    if(dep[u] > dep[v]) swap(u, v);
    for(int k = 0; k < 32; k ++) {
        if((dep[v] - dep[u]) >> k & 1)
            v = gfa[k][v];
    }
    if(u == v) return u;
    for(int k = 31; k >= 0; k --) {
        if(gfa[k][u] != gfa[k][v]) {
            u = gfa[k][u];
            v = gfa[k][v];
        }
    }
    return gfa[0][u];
}

int get_dis(int u, int v)
{
    return dis[u] + dis[v] - 2 * dis[LCA(u, v)];
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> t;
    while(t --) {
        cin >> n >> q;
        int ru, rv, rx;
        for(int i = 1; i <= n; i ++) {
            father[i] = i;
            G[i].clear();
        }
        for(int i = 0; i < n; i ++) {
            int a, b, x; cin >> a >> b >> x;
            int ra = get_root(a);
            int rb = get_root(b);
            if(ra == rb) {
                ru = a, rv = b, rx = x;
                continue;
            }
            father[ra] = rb;
            G[a].push_back(P(x, b));
            G[b].push_back(P(x, a));
        }
        init();
        while(q --) {
            int u, v; cin >> u >> v;
            int ans = get_dis(u, v);
            ans = min(ans, get_dis(ru, u) + rx + get_dis(rv, v));
            ans = min(ans, get_dis(ru, v) + rx + get_dis(rv, u));
            cout << ans << endl;
        }
    }
}

/*

The WAM is F**KING interesting .

*/

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