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只枚举起点到终点最短路上的所有边 看删除当前边后起点到终点最短距离是多少
因为删除最短路以外的边对于起点与终点距离没有影响 而最短路上最多（n-1）条边 时间复杂度为 (n^2)*(logn)
using namespace std;
struct nodeI
{
int v;
int w;
int next;
};
struct nodeII
{
friend bool operator < (nodeII n1,nodeII n2)
{
return n1.w>n2.w;
}
int v;
int w;
};
nodeI edge[1000010];
int first[1010],pre[1010],dis[1010],book[1010],flag[1000010];
int n,m,num,ans;
void addedge(int u,int v,int w)
{
edge[num].v=v;
edge[num].w=w;
edge[num].next=first[u];
first[u]=num++;
return;
}
void calculateI()
{
priority_queue <nodeII> que;
nodeII cur,tem;
int i,u,v,w;
memset(pre,-1,sizeof(pre));
memset(dis,0x3f,sizeof(dis));
memset(book,0,sizeof(book));
tem.v=1,tem.w=0;
que.push(tem);
dis[1]=0;
while(!que.empty())
{
cur=que.top();
que.pop();
u=cur.v;
if(book[u]==1) continue;
book[u]=1;
for(i=first[u];i!=-1;i=edge[i].next)
{
v=edge[i].v,w=edge[i].w;
if(book[v]==0&&dis[v]>dis[u]+w)
{
pre[v]=i;
dis[v]=dis[u]+w;
tem.v=v,tem.w=dis[v];
que.push(tem);
}
}
}
return;
}
void calculateII(int p)
{
priority_queue <nodeII> que;
nodeII cur,tem;
int i,u,v,w;
memset(dis,0x3f,sizeof(dis));
memset(book,0,sizeof(book));
tem.v=1,tem.w=0;
que.push(tem);
dis[1]=0;
while(!que.empty())
{
cur=que.top();
que.pop();
u=cur.v;
if(book[u]==1) continue;
book[u]=1;
for(i=first[u];i!=-1;i=edge[i].next)
{
if(i==p||(i^1)==p) continue;
v=edge[i].v,w=edge[i].w;
if(book[v]==0&&dis[v]>dis[u]+w)
{
pre[v]=i;
dis[v]=dis[u]+w;
tem.v=v,tem.w=dis[v];
que.push(tem);
}
}
}
ans=max(ans,dis[n]);
return;
}
void safari()
{
int i;
memset(flag,0,sizeof(flag));
for(i=pre[n];i!=-1;i=pre[edge[i^1].v])
{
flag[i]=1;
}
ans=0;
for(i=1;i<=m*2;i++)
{
if(flag[i])
{
calculateII(i);
}
}
return;
}
int main()
{
int i,u,v,w;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(first,-1,sizeof(first));
num=0;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
calculateI();
safari();
printf("%d\n",ans);
}
return 0;
}
