次最短路

次最短路的思想：
起点s, 终点t,中间点u,次最短路可以通过到u的最短路+u到t的次短路，或者t到u 的最短路+u到t的最短路。
例题：Roadblocks
代码：

#include<iostream>
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;

const int maxn = 200100;
const int INF = 1e9;
int h[maxn], to[maxn], w[maxn], nex[maxn], tot = 0;
void add(int u, int v, int vv){
     to[tot] = v;
     w[tot] = vv;
     nex[tot] = h[u];
     h[u] = tot++;
}

int dis[5010], dis1[5010];
int n, m;
typedef pair<int, int> PP;
int dijstra(){
    for(int i = 1; i <= n; i++){
        dis[i] = INF;
        dis1[i] = INF;
    }
    dis[1] = 0;
    priority_queue<PP, vector<PP>, greater<PP> >q;
    q.push({0, 1});
    while(!q.empty()){
        int u = q.top().second;
        int d = q.top().first;
        q.pop();
        if(dis1[u] < d)continue;

        for(int i = h[u]; i != -1; i = nex[i]){
            int val = w[i], v = to[i];
            int d2 = d + val;
            if(d2 < dis[v]){
                dis[v] = d2;
                q.push({d2, v});
            }
            if(d2 > dis[v] && d2 < dis1[v]){
                dis1[v] = d2;
                q.push({d2, v});
            }

        }

    }
    return dis1[n];
}
int main(){
    scanf("%d %d", &n, &m);
    memset(h, -1, sizeof h);
    for(int i = 1; i <= m; i++){
        int u, v, w;
        scanf("%d %d %d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }
    printf("%d\n", dijstra());
return 0;
}