BZOJ 2100: [Usaco2010 Dec]Apple Delivery

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2100

解：

典型的最短路，从两个点都跑一遍模板就好了，不过有点麻烦的是，如果跑dijkstra，要用堆来优化，如果跑spfa要用SLF（当然习惯用LLL也行）来优化，总之不能裸的模板去跑。

程序：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define INF 2100000000
using namespace std;
struct ding{
  int to,w,next;
}edge[400010];
struct ding2{
  int p,di;
  bool operator<(const ding2&a) const
  {
      return a.di<di;
  }
};
int cnt,ans,n,m;
int head[100010],dis[100010];
priority_queue<ding2>q;
void add(int u,int v,int d){edge[++cnt].to=v; edge[cnt].w=d; edge[cnt].next=head[u];head[u]=cnt;}
void dijkstra(int st)
{
  for (int i=1;i<=n;i++) dis[i]=INF;
  q.push((ding2){st,0});
  dis[st]=0;
  while (!q.empty())
  {
    ding2 now=q.top(); q.pop();
    if (now.di!=dis[now.p]) continue;
    for (int i=head[now.p];i;i=edge[i].next)
    {
      int k=edge[i].to;
      if (dis[k]>now.di+edge[i].w)
      {
        dis[k]=now.di+edge[i].w;
        q.push((ding2){k,dis[k]});
      }
    }
  }
}
int main()
{
  int st,en1,en2;
  scanf("%d%d%d%d%d",&m,&n,&st,&en1,&en2);
  int x,y,d;
  for (int i=1;i<=m;i++)
  {
      scanf("%d%d%d",&x,&y,&d);
      add(x,y,d); add(y,x,d);
  }
  dijkstra(en1);
  ans+=dis[en2]+dis[st];
  dijkstra(en2);
  ans=min(ans,dis[en1]+dis[st]);
  printf("%d\n",ans);
  return 0;
}