Hdu3592 World Exhibition --- difference Constraint

Hdu3592 World Exhibition --- difference Constraint

There is nothing special about this question.

X conditions: Sb-Sa <= c

Y conditions: Sa-Sb <=-c

The question is the relationship between 1 and n.

If there is a negative ring, the entire process is impossible. Output-1

If the graph is connected (1 to n are connected), Output d [n]

Disconnections are the case where-2 is mentioned in the question.

In the past, we usually add an additional node for creating a graph, or we start to team all the nodes. This means that we can add a meaningless condition considering the problem of disconnectivity.

#include 
 
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        #include 
        #define inf 0x3f3f3f3f#define eps 1e-6#define ll __int64using namespace std;#define N 1010struct node{ int v,w,next;}e[30020];int d[N],inq[N],outq[N],n,head[N],h;void addedge(int a,int b,int c){ e[h].v=b; e[h].w=c; e[h].next=head[a]; head[a]=h++;}int spfa(int s){ memset(d,0x3f,sizeof d); memset(inq,0,sizeof inq); memset(outq,0,sizeof outq); d[s]=0;inq[s]=1; queue
         
           q; q.push(s); int i,x; while(!q.empty()) { x=q.front(); q.pop(); inq[x]=0; outq[x]++; if(outq[x]>n) return 0; for(i=head[x];i!=-1;i=e[i].next) { if(d[e[i].v]>d[x]+e[i].w) { d[e[i].v]=d[x]+e[i].w; if(!inq[e[i].v]) { inq[e[i].v]=1; q.push(e[i].v); } } } } return 1;}void init(){ memset(head,-1,sizeof head); h=0;}int main(){ int T,a,b,c,x,y; scanf("%d",&T); while(T--) { init(); scanf("%d%d%d",&n,&x,&y); while(x--) { scanf("%d%d%d",&a,&b,&c); addedge(a,b,c); } while(y--) { scanf("%d%d%d",&a,&b,&c); addedge(b,a,-c); } if(!spfa(1)) printf("-1\n"); else if(d[n]!=inf) printf("%d\n",d[n]); else printf("-2\n"); } return 0;}