题目意思

给出t组数据，每组数据第一行n个节点m条边的无向图
给出边权信息，问选其中三个点最短路径之和最大值为多少。

解题思路

既然要我们最短路可能要用到dijkstra，又因为点可以随意确定，所以只要枚举中间点k，对k求最短路里面最长的两条路之和在return主函数判断最大值即可。如果只有一条路记得return 0出去主函数。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 1e9 + 7;
const ll INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e3 + 7; //节点数
const int M = 1e3 + 7; //路径数
int head[N], tot = 0;//前向星变量
struct Node {
    //int u; //起点
    int w; //权值
    int v, next;
} edge[M << 1];

void add(int u, int v, int w) {
    tot++;
    //edge[tot].u = u;
    edge[tot].v = v;
    edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot;
}

int dis[N], vis[N];

int dijkstra(int s) {
    memset(vis, 0, sizeof vis);
    memset(dis, 127, sizeof dis);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>        pq;
    dis[s] = 0;
    pq.push({ 0,s });
    int x = 0, y = 0;
    while (pq.size()) {
        pair<int, int> now = pq.top(); pq.pop();
        int u = now.second;
        if (vis[u])    continue;
        vis[u] = 1;
        y = max(y, dis[u]);
        if (y > x)    swap(x, y);
        for (int i = head[u]; i; i = edge[i].next) {
            if (vis[edge[i].v])    continue;
            if (dis[edge[i].v] > dis[u] + edge[i].w) {
                dis[edge[i].v] = dis[u] + edge[i].w;
                pq.push({ dis[edge[i].v],edge[i].v });
            }
        }
    }
    if (!y)    return 0;
    return x + y;
}

int main() {
    int T = read();
    while (T--) {
        tot = 0;
        memset(head, 0, sizeof head);
        int n = read(), m = read();
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            int u = read(), v = read(), w = read();
            add(u, v, w), add(v, u, w);
        }
        for (int i = 1; i <= n; ++i)
            ans = max(ans, dijkstra(i));
        if (ans)    print(ans);
        else print(-1);
        putchar(10);
    }
    return 0;
}