One example teaches you how to be brave with the subject, and the example is brave

One example teaches you how to be brave with the subject, and the example is brave

I had A question before. I got stuck with 10 greedy items, but I was still suffering from 11th greedy items,

Won't a question be answered? Greedy. How can we be greedy!

At present, the data of NOIP questions is not clearly defined. It is simple. It is greedy for data!

 

-By greedy god CCL

 

 

 

 

 

Today I made a very bt question.

P3385 [TEMPLATE] Negative Ring

This is actually a wildcard search SPFA. As a wildcard search SPFA, I am not convinced: triumph:

Whoever gets stuck with me will be greedy! : Confused:

 

Try First

When judging the negative ring, there is a clear conclusion that when you go back to the start point after a wave occurs from the start point and find that the update can continue, there must be a negative ring!

I thought that the data it provided had multiple Unicom components, but it turned out that I thought too much. : Sweat_smile:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>using namespace std;const int MAXN=1e6+10;inline int read(){    char c=getchar();int f=1,x=0;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9')x=x*10+c-48,c=getchar();return x*f;}struct node{    int u,v,w,nxt;}edge[MAXN];int head[MAXN];int num=1;inline void add_edge(int x,int y,int z){    edge[num].u=x;    edge[num].v=y;    edge[num].w=z;    edge[num].nxt=head[x];    head[x]=num++;}int inger[MAXN];int vis[MAXN];int dis[MAXN];inline void pre(){    memset(head,-1,sizeof(head));    num=1;    memset(vis,0,sizeof(vis));    memset(dis,0x7f,sizeof(dis));}int n,m;inline void SPFA(){    queue<int>q;    q.push(1);    vis[1]=1;dis[1]=0;    while(q.size()!=0)    {        int p=q.front();        q.pop();        vis[p]=0;        for(int i=head[p];i!=-1;i=edge[i].nxt)        {            if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)            {                dis[edge[i].v]=dis[edge[i].u]+edge[i].w;                if(vis[edge[i].v]==0)                {                    vis[edge[i].v]=1;                    if(edge[i].v==1)                    {                        printf("YE5\n");                        return ;                    }                    q.push(edge[i].v);                }            }        }    }    printf("N0\n");}int main(){    int T=read();    while(T--)    {        pre();        n=read();m=read();        for(int i=1;i<=m;i++)        {            int x=read(),y=read(),z=read();            if(z<0)add_edge(x,y,z);            else add_edge(x,y,z),add_edge(y,x,z);        }        SPFA();    }    return 0;}

 

In this way, A can be used.

However, this seems to be a reference to the concept of dfs,

It gets stuck in the number of queues,

Under normal circumstances, a negative loop is generated only when the number of entries in a node exceeds n, but this is too slow,

We want to narrow down the n

 

Try Second

N is too big. What should I do?

Let's just random one: grin:

However .. : Dizzy_face:

 

Try Third

 

It seems that rand is not working.

Let's specify it.

Like this:

if(inger[edge[i].v]>=250){    printf("YE5\n");    return ;}

250

 

200

 

100

 

.... : Angry:

 

 

 

 

20

: Grinning:

 

Think about it,

SPFA time complexity: $ O (ke) $

$ K $ indicates the average number of teams, and $ e $ indicates the number of sides. The value of $ k $ is generally about 2. Therefore, if the number of times a node is added to the queue is greater than 20, a negative ring will appear.

Of course, if you go to the test room, you should write the dfs SPFA. After all, there will be only one chance at that time.