P1938 [USACO09NOV]找工就业Job Hunt

P1938 [USACO09NOV]找工就业Job Hunt
给边赋予价值，入边的权值为D-Ti,然后从起点开始跑最长路，如果钱的总数超过了D*C，也就是一定有一个城市走了两遍，则有正环，则输出-1

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<algorithm>
 #include<cmath>
 #include<ctime>
 #include<set>
 #include<map>
 #include<stack>
 #include<cstring>
 #define inf 2147483647
 #define ls rt<<1
 #define rs rt<<1|1
 #define lson ls,nl,mid,l,r
 #define rson rs,mid+1,nr,l,r
 #define N 100010
 #define For(i,a,b) for(int i=a;i<=b;i++)
 #define p(a) putchar(a)
 #define g() getchar()

 using namespace std;
 int D,P,C,F,S;
 int b[][];
 int x,y,v;
 int vis[];
 int ans;
 int d[];

 struct node{
     int n;
     int v;
     node *next;
 }*e[];

 queue<int>q;

 void in(int &x){
     int y=;
     char c=g();x=;
     while(c<''||c>''){
         if(c=='-')y=-;
         c=g();
     }
     while(c<=''&&c>=''){
         x=(x<<)+(x<<)+c-'';c=g();
     }
     x*=y;
 }
 void o(int x){
     if(x<){
         p('-');
         x=-x;
     }
     if(x>)o(x/);
     p(x%+'');
 }

 void push(int x,int y,int v){
     node *p;
     p=new node();
     p->n=y;
     p->v=v;
     if(e[x]==NULL)
         e[x]=p;
     else{
         p->next=e[x]->next;
         e[x]->next=p;
     }
 }

 void hd(){
     d[S]=D;
     q.push(S);
     while(!q.empty()){
         int t=q.front();
         q.pop();
         vis[t]=true;
         for(node *i=e[t];i;i=i->next){
             if(d[t]+i->v>d[i->n]){
                 d[i->n]=d[t]+i->v;
                 if(!vis[i->n])
                 q.push(i->n);
             }
         }
         vis[t]=false;
     }
 }

 int main(){
     in(D);in(P);in(C);in(F);in(S);
     For(i,,P){
         in(x);in(y);
         b[x][y]=;
     }
     For(i,,F){
         in(x);in(y);in(v);
         push(x,y,D-v);
         if(D-v!=)
             b[x][y]=D-v;
         else
             b[x][y]=;
     }
     For(i,,C)
         For(j,,C)
            if(b[i][j]==)
                push(i,j,D);
     For(i,,C)
         d[i]=-inf;
        hd();
        ans=-inf;
        For(i,,C)
            ans=max(ans,d[i]);
        if(ans>C*D)
            o(-);
        else
            o(ans);
     return ;
 }