次短路模板题吧

题意已经非常裸了：求无向图的1到n次短路。

直接套用最短路(dijkstra)的主要框架。但在这个的基础上添加另外一个数组dist2。

走到一条边的时候来三个判定：

1. dist[u] + weight < dist[v]直接可以更新最短路，新的次短路就是曾经的最短路。

2. dist[u] + weight > dist[v] && dist[u] + weight < dist2[v]不可更新最短路，但可以更新次短路。

3. dist2[u] + weight < dist2[v]可以更新次短路。

然后就可以差不多写出来了。

话说这道题数据水。

代码：

```cpp

include

include

include

const int maxn = 5005, maxm = 100005;

struct Edges

{

int next, to, weight;

} e[maxm << 1];

int head[maxn], tot;

int dist[maxn], dist2[maxn];

int n, m;

struct Heapnodes

{

int d, u;

bool operator < (const Heapnodes &rhs) const

{

return d > rhs.d;

}

};

void link(int u, int v, int w)

{

e[++tot] = (Edges){head[u], v, w};

head[u] = tot;

}

void dijkstra(int s, int t)

{

memset(dist, 0x3f, sizeof dist);

memset(dist2, 0x3f, sizeof dist2);// 0x7f???

std::priority_queue heap;

dist[s] = 0; heap.push((Heapnodes){dist[s], s});

while(!heap.empty())

{

Heapnodes x = heap.top(); heap.pop();

int d = x.d, u = x.u;

if(d != dist[u]) continue;

for(int i = head[u]; i; i = e[i].next)

{

int v = e[i].to;

if(dist[u] + e[i].weight < dist[v])

{

dist2[v] = dist[v];

dist[v] = dist[u] + e[i].weight;

heap.push((Heapnodes){dist[v], v});

}

if(dist[u] + e[i].weight > dist[v] && dist[u] + e[i].weight < dist2[v])

{

dist2[v] = dist[u] + e[i].weight;

heap.push((Heapnodes){dist[v], v});

}

if(dist2[u] + e[i].weight < dist2[v])

{

dist2[v] = dist2[u] + e[i].weight;

//heap.push((Heapnodes){dist[v], v});

}

}

}

}

int main()

{

scanf("%d%d", &n, &m);

while(m--)

{

int u, v, w; scanf("%d%d%d", &u, &v, &w);

link(u, v, w); link(v, u, w);

}

dijkstra(1, n);

printf("%d\n", dist2[n]);

return 0;

}

``