题目链接：https://www.luogu.com.cn/problem/P1821
题目大意：求所点 (i) 的从起点出发到 (i) 然后再从 (i) 返回起点的最短路径的最大值。
解题思路：
求起点到所有点的最短路可以直接得到。
求所有点到起点的最短路，可以建反图，然后用最短路算法求。
所以这道题相当于求两边最短路，其中第二遍是建反图求最短路。
实现代码如下：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010, maxm = 200010;
struct Edge {
int v, w, nxt;
Edge() {};
Edge(int _v, int _w, int _nxt) { v = _v; w = _w; nxt = _nxt; }
} edge[maxm];
int n, m, head[maxn], ecnt;
void init() {
ecnt = 0;
memset(head, -1, sizeof(int)*(n+1));
}
void addedge(int u, int v, int w) {
edge[ecnt] = Edge(v, w, head[u]); head[u] = ecnt ++;
}
struct Node {
int u, dis;
Node() {};
Node(int _u, int _dis) { u = _u; dis = _dis; }
bool operator < (const Node& x) const {
return dis > x.dis;
}
};
priority_queue<Node> que;
int dis[maxn];
bool vis[maxn];
void dijkstra(int s) {
for (int i = 1; i <= n; i ++) {
dis[i] = -1;
vis[i] = false;
}
while (!que.empty()) que.pop();
dis[s] = 0;
que.push(Node(s, 0));
while (!que.empty()) {
Node nd = que.top();
que.pop();
int u = nd.u;
if (vis[u]) continue;
vis[u] = true;
for (int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v, w = edge[i].w;
if (!vis[v] && (dis[v] == -1 || dis[v] > dis[u] + w)) {
dis[v] = dis[u] + w;
que.push(Node(v, dis[v]));
}
}
}
}
int s, u[maxm], v[maxm], w[maxm];
int dis1[maxn], dis2[maxn];
int main() {
cin >> n >> m >> s;
for (int i = 0; i < m; i ++) cin >> u[i] >> v[i] >> w[i];
init();
for (int i = 0; i < m; i ++) addedge(u[i], v[i], w[i]);
dijkstra(s);
for (int i = 1; i <= n; i ++) dis1[i] = dis[i];
init();
for (int i = 0; i < m; i ++) addedge(v[i], u[i], w[i]);
dijkstra(s);
for (int i = 1; i <= n; i ++) dis2[i] = dis[i];
int ans = 0;
for (int i = 1; i <= n; i ++) if (dis1[i] != -1 && dis2[i] != -1) ans = max(ans, dis1[i] + dis2[i]);
cout << ans << endl;
return 0;
}