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 ;

 }