題意：第一行是點數N和邊數R，後面是R條無向邊。求從1到n的次短路長度。

思路：模板題。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <cstdlib>
#include <set>
#include <string>

using namespace std;

typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 5005;
struct edge{
    int v, w, nxt;
}G[200005]; // 無向邊陣列要開兩倍
int tot, pre[maxn];
void add(int u, int v, int w) {
    G[tot].v = v;
    G[tot].w = w;
    G[tot].nxt = pre[u];
    pre[u] = tot++;
}
int dist[maxn][2], n, r;
bool vis[maxn][2];
struct node{
    int v, d, flag;
    node() {}
    node(int v, int d, int f) : v(v), d(d), flag(f) {}
    bool operator < (const node &b) const {
        if (d != b.d) {
            return d > b.d;
        } else {
            return v > b.v;
        }
    }
};
void dijkstra(int s) {
    memset(dist, 0x3f, sizeof(dist));
    memset(vis, 0, sizeof(vis));
    dist[s][0] = 0;
    priority_queue<node> que;
    que.push(node(s, 0, 0));
    while (!que.empty()) {
        node p = que.top();
        que.pop();
        if (vis[p.v][p.flag]) {
            continue;
        }
        vis[p.v][p.flag] = true;
        for (int i = pre[p.v]; ~i; i = G[i].nxt) {
            int v = G[i].v;
            int w = G[i].w;
            if (!vis[v][0] && p.d + w < dist[v][0]) {
                if (dist[v][0] != inf) {
                    dist[v][1] = dist[v][0];
                    que.push(node(v, dist[v][1], 1));
                }
                dist[v][0] = p.d + w;
                que.push(node(v, dist[v][0], 0));
            } else if (!vis[v][1] && p.d + w < dist[v][1]) {
                dist[v][1] = p.d + w;
                que.push(node(v, dist[v][1], 1));
            }
        }
    }
}
int main(){
    int u, v, w;
    scanf("%d%d", &n, &r);
    tot = 0;
    memset(pre, -1, sizeof(pre));
    while (r--) {
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }
    dijkstra(1);
    printf("%d\n", dist[n][1]);
    return 0;
}