POJ 3268 最短路应用

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
const int maxn=1000+1,maxm=100000+1,INF=1e+8;
int len[maxn][maxn],len1[maxn][maxn],dist[maxn],dist1[maxn];
vector<int> path[maxn],path1[maxn];
int n,m,source;
void SPFA(int dist[],int d[][maxn],vector<int>son[],int source)
{
 bool inq[maxn];
 int times[maxn];
 memset(inq,0,sizeof(inq));
 memset(times,0,sizeof(times));
 for(int i=1;i<=n;i++)
 	dist[i]=INF;
 dist[source]=0;
 queue<int>q;
 q.push(source);
 times[source]++;
 inq[source]=true;
 while(!q.empty())
 	{
 	 int now=q.front(); 
	 q.pop();
	 inq[now]=false;
	 for(int i=0;i<son[now].size();i++)
	 	{
	 	 int np=son[now][i];
	 	 if(dist[np]>dist[now]+d[now][np])
	 	 	{
	 	 	 dist[np]=dist[now]+d[now][np];
	 	 	 
	 	 	 if(times[np]>n)return;
	 	 	 if(!inq[np]){times[np]++;q.push(np);}
			}
		}	
	}
}

int main()
{freopen("t.txt","r",stdin);
 scanf("%d%d%d",&n,&m,&source);
 for(int i=0;i<m;i++)
 	{
 	 int a,b,l;
 	 scanf("%d%d%d",&a,&b,&l);
 	 path[a].push_back(b);len[a][b]=l;
 	 path1[b].push_back(a);len1[b][a]=l;
	}
 SPFA(dist,len,path,source);
 SPFA(dist1,len1,path1,source);
 int ans=0;
 for(int i=1;i<=n;i++)
 	{
 	 if((dist[i]+dist1[i])>ans)ans=dist[i]+dist1[i];
	}
 printf("%d\n",ans);
 return 0;
}