BZOJ 1631: [Usaco2007 Feb]Cow Party

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1631

解：先跑一遍最短路，把所有边都反过来以后，再跑一遍

程序：

#include
#include
#include
#include
#include
#define INF 2100000000
using namespace std;
int n,root,m,dis[2000],ans[2000];
bool vis[2000];
vectorint,int> >edge1[2000],edge2[2000];
void djk1()
{
  for (int i=1;i<=n;i++) {dis[i]=INF;vis[i]=false;}
  dis[root]=0; 
  for (;;)
  {
      int minx=INF,ch=-1;
      for (int i=1;i<=n;i++)
      if ((!vis[i])&&(dis[i]<minx))
      {
        minx=dis[i];
        ch=i;
    }
    if (ch==-1) return;
    vis[ch]=true;
    for (int i=0;i)
    {
      pair<int,int> y=edge1[ch][i];
      if (dis[y.first]>dis[ch]+y.second) dis[y.first]=dis[ch]+y.second;
    }
  }
}
void djk2()
{
  for (int i=1;i<=n;i++) {dis[i]=INF;vis[i]=false;}
  dis[root]=0; 
  for (;;)
  {
      int minx=INF,ch=-1;
      for (int i=1;i<=n;i++)
      if ((!vis[i])&&(dis[i]<minx))
      {
        minx=dis[i];
        ch=i;
    }
    if (ch==-1) break;
    vis[ch]=true;
    for (int i=0;i)
    {
      pair<int,int> y=edge2[ch][i];
      if (dis[y.first]>dis[ch]+y.second) dis[y.first]=dis[ch]+y.second;
    }
  }
}
int main()
{
  scanf("%d%d%d",&n,&m,&root);
  int x,y,t;
  for (int i=1;i<=m;i++)
  {
      scanf("%d%d%d",&x,&y,&t);
      edge1[x].push_back(make_pair(y,t));
      edge2[y].push_back(make_pair(x,t));
  }
  djk1();
  for (int i=1;i<=n;i++) ans[i]=dis[i];
  djk2(); 
  int now=0;
  for (int i=1;i<=n;i++) if (ans[i]+dis[i]>now) now=ans[i]+dis[i];
  printf("%d\n",now);
  return 0;
}

 

转载于:https://www.cnblogs.com/2014nhc/p/6660394.html