POJ No.3255 Roadblocks 求次短路径

 #define _CRT_SECURE_NO_WARNINGS
 /*
 7 10
 0 1 5
 0 2 2
 1 2 4
 1 3 2
 2 3 6
 2 4 10
 3 5 1
 4 5 3
 4 6 5
 5 6 9

 4 4
 0 1 100
 1 3 200
 1 2 250
 2 3 100
 */
 #include <iostream>
 #include <functional>
 #include <utility>
 #include <queue>
 #include <vector>
 using namespace std;

 const int maxn =  + ;
 const int INF = ;
 typedef pair<int, int> P;     //first-最短距离，second-顶点编号
 struct edge
 {
     int to, cost;
 };

 //输入
 int N, R;             //N-点，R-边
 vector<edge> G[maxn]; //图的邻接表表示

 int dist[maxn];   //最短距离
 int dist2[maxn];  //次短距离

 void solve()
 {
     priority_queue<P, vector<P>, greater<P> > que;
     fill(dist, dist + N, INF);
     fill(dist2, dist2 + N, INF);
     dist[] = ;
     que.push(P(, ));

     while (!que.empty())
     {
         P p = que.top(); que.pop();       //队列存放 （最短路和次短路）
         int v = p.second, d = p.first;
         if (dist2[v] < d) continue;       //如果次小的值小，跳过该值
         for (int i = ; i < G[v].size(); i++) {
             edge &e = G[v][i];            //得到v的临边
             int d2 = d + e.cost;          //d2 = 已知最短或次短 + 临边权值
             if (dist[e.to] > d2) {        //如果未确定的最短路 > d2
                 swap(dist[e.to], d2);
                 que.push(P(dist[e.to], e.to));   //添加最短路
             }
             if (dist2[e.to] > d2 && dist[e.to] < d2) {
                 dist2[e.to] = d2;                //说明d2是次大值
                 que.push(P(dist2[e.to], e.to));  //添加次短路
             }
         }
     }
     printf("%d\n", dist2[N - ]);        //N位置的dist2即为
 }

 void input()
 {
     int from, to, cost;
     edge tmp;
     for (int i = ; i < R; i++) {
         cin >> from >> to >> cost;
         tmp.to = to; tmp.cost = cost;
         G[from].push_back(tmp);
         tmp.to = from;
         G[to].push_back(tmp);
     }
 }

 int main()
 {
     cin >> N >> R;
     input();
     solve();
     return ;
 }