LightOJ 1074 Extended Traffic SPFA 消负环

Posted 2020-06-29 shuguangzw

分析：一看就是求最短路，然后用dij，果断错了一发，发现是3次方，有可能会出现负环

然后用spfa判负环，然后标记负环所有可达的点，被标记的点答案都是“？”

#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
typedef long long LL;
const int N=200+5;
const int INF=0x7fffffff;
struct Edge{
   int v;
   LL w;
   int next;
   bool operator<(const Edge &e)const{
      return w>e.w;
   } 
}edge[N*N];
int head[N],val[N],tot,n,m,p;
LL d[N];
void add(int u,int v,LL w){
   edge[tot].v=v;
   edge[tot].w=w;
   edge[tot].next=head[u];
   head[u]=tot++;
}
bool inq[N];
queue<int>q;
int cnt[N];
bool c[N];
void dfs(int u){
   c[u]=1;
   for(int i=head[u];~i;i=edge[i].next)
     if(!c[edge[i].v])dfs(edge[i].v);
}
void spfa(int s){
   for(int i=1;i<=n;++i)
     c[i]=cnt[i]=inq[i]=0,d[i]=INF;
   q.push(s),cnt[s]=1,inq[s]=1,d[s]=0;
   while(!q.empty()){
     int u=q.front();
     q.pop();
     inq[u]=0;
     if(c[u])continue;
     for(int i=head[u];~i;i=edge[i].next){
         int v=edge[i].v;
         if(c[v])continue;
         if(d[v]>d[u]+edge[i].w){
           d[v]=d[u]+edge[i].w;
           if(inq[v])continue;
            q.push(v);
            ++cnt[v];
            inq[v]=1;
            if(cnt[v]>n)dfs(v);
         }
     }
   }
}
int main(){ 
   int T,cas=0;
   scanf("%d",&T);
   while(T--){
     scanf("%d",&n);
     for(int i=1;i<=n;++i)scanf("%d",&val[i]);
      scanf("%d",&m);
      memset(head,-1,sizeof(head)),tot=0;
      for(int i=0;i<m;++i){
        int u,v;
        scanf("%d%d",&u,&v);
        LL w=(val[v]-val[u]);
        w=w*w*w;
        add(u,v,w);
      }
      spfa(1);
      scanf("%d",&p);
      printf("Case %d:\n",++cas);
      for(int i=1;i<=p;++i){
         int u;
         scanf("%d",&u);
         if(c[u]||d[u]==INF||d[u]<3)printf("?\n");
         else printf("%d\n",d[u]);
      }
   }
    return 0;
}

View Code