poj3255 次短路的长度

　　这道问题是求1-N的次短路的长度，我们直接在dist[maxn][2]上加1维更新即可， 代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;
const int maxn = 5000 + 10;

int N, R;
struct edge { int v, c; };
vector<edge> G[maxn];

struct Dij
{
    int u, flog, c;
    bool operator< (const Dij& r) const
    {
        return c > r.c;
    }
};

int dist[maxn][2], vis[maxn][2];
void dijkstra()
{
    memset(dist, 0x3f, sizeof(dist));
    memset(vis, 0, sizeof(vis));
    dist[1][0] = 0;
    priority_queue<Dij> que;
    que.push((Dij){1, 0, 0});
    while(!que.empty())
    {
        Dij tp = que.top(); que.pop();
        int u=tp.u, flog=tp.flog;
        if(vis[u][flog]) continue;
        vis[u][flog] = 1;
        for(int i=0; i<G[u].size(); i++)
        {
            int v = G[u][i].v, c = G[u][i].c;
            int w = dist[u][flog] + c;
            if(w < dist[v][0])    //更新次短路 最短路
            {
                if(dist[v][0] != 0x3f3f3f3f)
                {
                    dist[v][1] = dist[v][0];
                    que.push((Dij){v, 1, dist[v][1]});
                }
                dist[v][0] = w;
                que.push((Dij){v, 0, dist[v][0]});
            }
            else if(w<dist[v][1]) //更新次短路
            {
                dist[v][1] = w;
                que.push((Dij){v, 1, dist[v][1]});
            }
        }
    }
}

int main()
{
    scanf("%d%d", &N, &R);
    for(int i=0; i<R; i++)
    {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        G[u].push_back((edge){v, c});
        G[v].push_back((edge){u, c});
    }
    dijkstra();
    printf("%d\n", dist[N][1]);
    return 0;
}