Welcome to the Hungary Games! The streets of Budapest form a twisted network of one-way streets.
欢迎来到匈牙利游戏！布达佩斯（匈牙利首都）的街道形成了一个弯曲的单向网络。
You have been forced to join a race as part of a “Reality TV” show where you race through these streets, starting at the Sz´echenyi thermal bath (s for short) and ending at the Tomb of G¨ ul Baba (t for short).
你被强制要求参加一个赛跑作为一个TV秀的一部分节目，比赛中你需要穿越这些街道，从s开始，到t结束。
Naturally, you want to complete the race as quickly as possible, because you will get more promo- tional contracts the better you perform.
很自然的，你想要尽快的完成比赛，因为你的比赛完成的越好，你就能得到更多的商业促销合同。
However, there is a catch: any person who is smart enough to take a shortest s-t route will be thrown into the P´alv¨olgyi cave system and kept as a national treasure. You would like to avoid this fate, but still be as fast as possible. Write a program that computes a strictly-second-shortest s-t route.
但是，有一个需要了解的是，如果有人过于聪明找到从s到t的最短路线，那么他就被扔到国家*人类保护系统中作为一个国家宝藏收藏起来。你显然要避免这种事情的发生，但是也想越快越好。写一个程序来计算一个从s到t的严格次短路线吧。
Sometimes the strictly-second-shortest route visits some nodes more than once; see Sample Input 2 for an example.
有的时候，严格次短路线可能访问某些节点不止一次。样例2是一个例子。
The first line will have the format N M, where N is the number of nodes in Budapest and M is the number of edges. The nodes are 1,2,...,N; node 1 represents s; node N represents t. Then there are M lines of the form A B L, indicating a one-way street from A to B of length L. You can assume that A != B on these lines, and that the ordered pairs (A,B) are distinct.
第一行包含两个整数N和M，N代表布达佩斯的节点个数，M代表边的个数。节点编号从1到N。1代表出发点s，N代表终点t。接下来的M行每行三个整数A B L，代表有一条从A到B的长度为L的单向同路。你可以认为A不等于B，也不会有重复的(A,B)对。
Output the length of a strictly-second-shortest route from s to t. If there are less than two possible lengths for routes from s to t, output −1.
输出从s到t的严格次短路的长度。如果从s到t的路少于2条，输出-1。
样例输入1：
4 6
1 2 5
1 3 5
2 3 1
2 4 5
3 4 5
1 4 13
样例输入2：
2 2
1 2 1
2 1 1
样例输出1：
11
样例输出2：
3
对于样例1：
There are two shortest routes of length 10 (1 → 2 → 4,1 → 3 → 4) and the strictly-second- shortest route is 1 → 2 → 3 → 4 with length 11.
对于样例2：
The shortest route is 1 → 2 of length 1, and the strictly-second route is 1 → 2 → 1 → 2 of length 3.
严格次短路、
#include <cstring>
#include <cstdio>
#include <queue> const int N(+);
const int M(+);
int n,hed[N],had[N],sumedge;
struct Edge
{
int v,next,w;
Edge(int v=,int next=,int w=):
v(v),next(next),w(w){}
}edge1[M],edge2[M];
inline void ins(int u,int v,int w)
{
edge1[++sumedge]=Edge(v,hed[u],w);
hed[u]=sumedge;
edge2[sumedge]=Edge(u,had[v],w);
had[v]=sumedge;
} bool inq[N];
int d1[N],d2[N];
void SPFA(int s,int *dis,int *head,Edge *edge)
{
std::queue<int>que;
for(int i=;i<=n;i++)
dis[i]=0x3f3f3f3f,inq[i]=;
dis[s]=; que.push(s);
for(int u,v;!que.empty();)
{
u=que.front(); que.pop(); inq[u]=;
for(int i=head[u];i;i=edge[i].next)
{
v=edge[i].v;
if(dis[v]>dis[u]+edge[i].w)
{
dis[v]=dis[u]+edge[i].w;
if(!inq[v]) inq[v]=,que.push(v);
}
}
}
} inline void read(int &x)
{
x=; register char ch=getchar();
for(;ch>''||ch<'';) ch=getchar();
for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-'';
} int AC()
{
int m;
read(n),read(m);
for(int u,v,w;m--;ins(u,v,w))
read(u),read(v),read(w);
SPFA(,d1,hed,edge1);
SPFA(n,d2,had,edge2);
int ans=0x3f3f3f3f;
for(int u=;u<=n;u++)
{
for(int v,i=hed[u];i;i=edge1[i].next)
{
v=edge1[i].v;
int tmp=d1[u]+d2[v]+edge1[i].w;
if(tmp<ans&&tmp>d1[n]) ans=tmp;
}
}
if(ans==0x3f3f3f3f) ans=-;
printf("%d\n",ans);
return ;
} int Hope=AC();
int main(){;}
AC的SPFA
下面可以忽略、
#include <cstring>
#include <cstdio>
#include <queue> const int N(+);
const int M(+);
int hed[N],had[N],sumedge;
struct Edge
{
int v,next,w;
Edge(int v=,int next=,int w=):
v(v),next(next),w(w){}
}edge1[M],edge2[M];
inline void ins(int u,int v,int w)
{
edge1[++sumedge]=Edge(v,hed[u],w);
hed[u]=sumedge;
edge2[sumedge]=Edge(u,had[v],w);
had[v]=sumedge;
} int dis[N];
bool inq[N];
void SPFA(int s)
{
for(int i=;i<=s;i++) dis[i]=0x3f3f3f3f;
std::queue<int>que;
dis[s]=; que.push(s);
for(int u,v;!que.empty();)
{
u=que.front(); que.pop(); inq[u]=;
for(int i=had[u];i;i=edge2[i].next)
{
v=edge2[i].v;
if(dis[v]>dis[u]+edge2[i].w)
{
dis[v]=dis[u]+edge2[i].w;
if(!inq[v]) que.push(v),inq[v]=;
}
}
}
} struct Node
{
int now,g;
bool operator < (const Node x)const
{
return dis[now]+g>dis[x.now]+x.g;
}
};
int A_star(int s,int t,int k)
{
if(dis[s]==0x3f3f3f3f) return -;
int ret=0x3f3f3f3f,cnt=;
std::priority_queue<Node>que;
Node u,v; u.now=s; u.g=;
que.push(u);
for(;!que.empty();)
{
u=que.top(); que.pop();
if(cnt>k) return -;
if(u.now==t)
{
if(ret!=u.g)
{
ret=u.g;
if(++cnt==)
return ret;
}
}
for(int i=hed[u.now];i;i=edge1[i].next)
{
v.now=edge1[i].v;
v.g=u.g+edge1[i].w;
que.push(v);
}
}
return -;
} inline void read(int &x)
{
x=; register char ch=getchar();
for(;ch>''||ch<'';) ch=getchar();
for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-'';
} int AC()
{
int n,m,k;
read(n),read(m);k=m;
for(int u,v,w;m--;ins(u,v,w))
read(u),read(v),read(w);
SPFA(n);
printf("%d\n",A_star(,n,k));
return ;
} int Hope=AC();
int main(){;}
爆栈的Astar
#include <cstdio>
#include <queue> using namespace std; const int INF(0x3f3f3f3f);
const int N(+);
const int M(+); int m,n,head[N],sumedge;
struct Edge
{
int v,next;
long long w;
Edge(int v=,int next=,long long w=):
v(v),next(next),w(w){}
}edge[M<<];
inline void ins(int u,int v,long long w)
{
edge[++sumedge]=Edge(v,head[u],w);
head[u]=sumedge;
} struct Node
{
int now;
long long dis;
bool operator < (const Node &x) const
{
return dis>x.dis;
}
}; bool vis[N];
long long d1[N],d2[N];
#define swap(a,b) {long long tmp=a;a=b;b=tmp;}
inline void Dijkstar()
{
for(int i=;i<=n;i++) d1[i]=d2[i]=INF;
priority_queue<Node>que; Node u,to;
u.dis=d1[]=; vis[]=;
u.now=; que.push(u);
for(int dis,v;!que.empty();)
{
u=que.top();que.pop();
if(u.dis>d2[u.now]) continue;
for(int i=head[u.now];i;i=edge[i].next)
{
v=edge[i].v;
dis=u.dis+edge[i].w;
if(dis<d1[v])
{
swap(dis,d1[v]);
to.now=v;
to.dis=d1[v];
que.push(to);
}
if(dis>d1[v]&&dis<d2[v])
{
d2[v]=dis;
to.dis=d2[v];
to.now=v;
que.push(to);
}
}
}
} inline void read(int &x)
{
x=; register char ch=getchar();
for(;ch>''||ch<'';) ch=getchar();
for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-'';
} int AC()
{ read(n),read(m);
for(int v,u,w;m--;ins(u,v,(long long)w))
read(u),read(v),read(w);
Dijkstar();
if(d2[n]==d1[n]) d2[n]=-;
printf("%lld\n",d2[n]);
return ;
} int I_want_AC=AC();
int main(){;}
不能输出-1的Dijkstra