题意：给你一张简单无向图，问你1到n的次短路。注意，可以不是简单路径。

存个次短路板子，原理还是挺简单，直接看代码吧。然后这份代码还是个fread的示例用法。

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const int BUF=60000000;char Buf[BUF],*buf=Buf;inline void read(int&a){for(a=0;*buf<48;buf++);while(*buf>47)a=a*10+*buf++-48;}#define N 100000+10#define INF 1000000000000000lltypedef long long ll;typedef pair
         
           Point;priority_queue
          
           
            ,greater
            
              >Heap;int n;ll dist[N],dist2[N];int e,first[N],nex[N<<1],v[N<<1],w[N<<1];void AddEdge(int U, int V,int W){ v[++e]=V; w[e]=W; nex[e]=first[U]; first[U]=e;}int m,T;int main(){ int x,y,z;// freopen("1011.in","r",stdin); fread(Buf,1,BUF,stdin); read(T); for(;T;--T){ read(n); read(m); e=0; memset(first,0,sizeof(first)); for(int i=1;i<=m;++i){ read(x); read(y); read(z); AddEdge(x-1,y-1,z); AddEdge(y-1,x-1,z); } fill(dist,dist+n,INF); fill(dist2,dist2+n,INF); dist[0]=0; Heap.push(Point(0,0)); while(!Heap.empty()){ Point o=Heap.top(); Heap.pop(); int U=o.second; ll d=o.first; if(dist2[U]
             
              d2){ swap(dist[v[i]],d2); Heap.push(Point(dist[v[i]],v[i])); } if(dist2[v[i]]>d2 && dist[v[i]]