P1993 small farm problem solution of K

Luo Gu original page

solution:

This is a differential constraint problem is not constrained differential system of human way, please
differential restraint system in the form of a triangle inequality much like single-source shortest path, it has been listed in graph theory algorithms to equation (I was seeking at least by worth seeking algorithm, find the maximum contrast)

\[ \left\{\begin{matrix} & a-b \geq c(k=1)& \\ & b-a \geq c (k=2)& \\ & a-b \geq 0, b-a \geq 0(k=3) & \end{matrix}\right.\]

If the construction of FIG negative ring (longest run is seeking ring road), i.e., small KSmokedmisremembered.

I had to fight this problem is based on BFS_SPFA, then overtime,SPFA has died, feelI changed dishesAlthough DFS_SPFA easy to use but still exponential complexity,In particular there is likely to be the topic of people jammed cancerTLE is easy, because it is subject data range \ (n, m \ leq 10000 \) I can be divided into two in the embodiment, if \ (n-m * \. 7 GEQ 1E + \) , then hung DFS_SPFA, if not meet, selectKe learning methodBFS_SPFA,

BFS_SPFA find negative ring

An array as referred to \ (CNT [I] \) , represents the number of sides to, if \ (CNT [I] \ GEQ m \) , i.e., negative loop, high efficiency of this methodBut the worst is O (nm) or deadThan many online sacrifice correctness to run more harm than good and sometimes better algorithms.

CodeI know that you like the most：

#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#define re register
using namespace std;
template<typename T>
inline void read(T&x)
{
    x=0;
    char s=getchar();
    bool f=false;
    while(!(s>='0'&&s<='9'))
    {
        if(s=='-')
            f=true;
        s=getchar();
    }
    while(s>='0'&&s<='9')
    {
        x=(x<<1)+(x<<3)+s-'0';
        s=getchar();
    }
    if(f)
        x=(~x)+1;
}
const int N=1000010;
int dis[N];
struct Edge
{
    int next,to,dis;
} edge[N];
int num_edge,head[N],n,m,cnt[N],*v,*w;
bool exist[N];
inline void add_edge(const int&from,const int&to,const int&dis)
{
    edge[++num_edge].next=head[from];
    head[from]=num_edge;
    edge[num_edge].to=to;
    edge[num_edge].dis=dis;
}
inline void dfs_spfa(int u)
{
    exist[u]=true;
    for(re int i=head[u]; i; i=edge[i].next)
    {
        v=&edge[i].to;
        w=&edge[i].dis;
        if(dis[*v]<dis[u]+(*w))
        {
            dis[*v]=dis[u]+(*w);
            if(exist[*v])
            {
                printf("No\n");
                exit(0);
            }
            else
                dfs_spfa(*v);
        }
    }
    exist[u]=false;
}
queue<int>q;
inline void bfs_spfa()
{
    dis[0]=0;
    q.push(0);
    re int u;
    do
    {
        u=q.front();
        q.pop();
        exist[u]=false;
        for(re int i=head[u]; i; i=edge[i].next)
        {
            v=&edge[i].to;
            w=&edge[i].dis;
            if(dis[*v]<dis[u]+(*w))
            {
                dis[*v]=dis[u]+(*w);
                cnt[*v]=cnt[u]+1;
                if(cnt[*v]>m)
                {
                    printf("No\n");
                    return;
                }
                if(!exist[*v])
                {
                    exist[*v]=true;
                    q.push(*v);
                }
            }
        }
    }
    while(!q.empty());
    printf("Yes\n");
}
int main()
{
    read(n);
    for(re int i=1; i<=n; i++)
        dis[i]=-0x3f3f3f3f;
    read(m);
    for(re int i=1,k,a,b,c; i<=m; i++)
    {
        read(k);
        read(a);
        read(b);
        if(k==1)
        {
            read(c);
            add_edge(b,a,c);
        }
        else if(k==2)
        {
            read(c);
            add_edge(a,b,-c);
        }
        else if(k==3)
        {
            add_edge(a,b,0);
            add_edge(b,a,0);
        }
    }
    for(re int i=1; i<=n; i++)
        add_edge(0,i,0);
    if(n*m>=1e7)
    {
        dfs_spfa(0);
        printf("Yes\n");
    }
    else
        bfs_spfa();
    return 0;
}

If the differential restraint system has been a great foundation, it is recommended to do
[SCOI2011] candy
and cashier issues

Thank you for watching! ! !