Poj 1511 invitation cards

Source: http://poj.org/problem? Id = 1511

It takes a long time to understand the question. The goal is to find the sum of the two minimum paths (that is, the travel expenses of this question,

For the first time, the path from CCS (1) to each point is the smallest. The spfa algorithm does not have to say that when it comes back, the end point is determined by CCS (1), which is equivalent to putting the path

In turn, the digraph goes in the inverse direction, and the path from 1 to each point is the smallest, and then a spfa is used. Note that ANS must use long.

Otherwise, it would also be wa, Where wa was made several times. Even though the AC was changed, I still don't understand it. The question clearly says smaller than 1000000000,

It's speechless.

 1 #include<stdio.h> 2 #include<string.h> 3 #define INF 0x3f3f3f3f 4 const int maxn=1000000+10; 5 int u[maxn],v[maxn],w[maxn],next[maxn],first[maxn],dist[maxn],que[maxn]; 6 bool inq[maxn]; 7 void add_edge(int a,int e) 8 { 9     next[e]=first[a];10     first[a]=e;11 }12 void spfa(void)13 {14     int head,tail;15     dist[1]=0;16     que[head=tail=0]=1;17     tail++;18     inq[1]=true;19     while(head!=tail){20         int a=que[head];21         head=(head+1)%maxn;22         inq[a]=false;23         for(int e=first[a];e!=-1;e=next[e]){24             if(dist[v[e]]>dist[a]+w[e]){25                 dist[v[e]]=dist[a]+w[e];26                 if(!inq[v[e]]){27                     inq[v[e]]=true;28                     que[tail]=v[e];29                     tail=(tail+1)%maxn;30                 }31             }32         }33     }34 }35 int main()36 {37     int t,q,p;38     scanf("%d",&t);39     while(t--){40         scanf("%d%d",&p,&q);41         memset(first,-1,sizeof(int)*(p+1));42         for(int e=1;e<=q;e++){43             scanf("%d%d%d",&u[e],&v[e],&w[e]);44             add_edge(u[e],e);45         }46         memset(dist,0x3f,sizeof(int)*(p+1));47         memset(inq,false,sizeof(bool)*(p+1));48         spfa();49         long long ans=0;50         for(int i=1;i<=p;i++)51             ans+=dist[i];52         memset(first,-1,sizeof(int)*(p+1));53         for(int e=1;e<=q;e++){54             int t=u[e];55             u[e]=v[e];56             v[e]=t;57             add_edge(u[e],e);58         }59         memset(dist,0x3f,sizeof(int)*(p+1));60         memset(inq,false,sizeof(bool)*(p+1));61         spfa();62         for(int i=1;i<=p;i++)63             ans+=dist[i];64         printf("%lld\n",ans);65     }66     return 0;67 }