P1993 small K farm, p1993 farm

P1993 small K farm, p1993 farm
Description

Little K builds many farms in Minecraft, n in total, so that he forgets Every

The number of crops planted on the farm. He only remembers some vague information (m in total ).

Description:

However, because the memory of Xiao K is somewhat biased, he wants to know that there is no such situation

The number of crops is consistent with all the information in his memory.

Input/Output Format

Input Format:

 

Input data from farm. in

The first line contains two integers, n and m, indicating the number of farms and the number of small K memories.

Next, m rows:

If the first number of each row is 1, there are three integers a, B, c, indicating that farm a is planted more than farm B.

A crop of c units.

If the first number of each row is 2, there are three integers a, B, c, indicating that farm a is more planted than farm B.

A crop of c units. If the first number in each row is 3, there are two integers, a and B, indicating that farm a is terminated.

The number is the same as that of B.

 

Output Format:

 

Output to farm. out

If there is a situation that matches the memory of small K, "Yes" is output; otherwise, "No" is output ".

 

Input and Output sample input sample #1:

3 33 1 21 1 3 12 2 3 2

Output sample #1:

Yes

Description

100% data guarantee: 1 ≤ n, m, a, B, c ≤ 10000.

 

Remember to use deep search to write the difference constraint in the future, otherwise it is difficult to judge the negative ring.

 

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 const int MAXN=100001; 9 const int maxn=0x7ffffff;10 void read(int &n)11 {12     char c='+';int x=0;bool flag=0;13     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}14     while(c>='0'&&c<='9'){x=x*10+c-48;c=getchar();}15     flag==1?n=-x:n=x;16 }17 int n,m;18 struct node19 {20     int u,v,w,nxt;21 }edge[MAXN];22 int head[MAXN];23 int num=1;24 void add_edge(int x,int y,int z)25 {26     edge[num].u=x;27     edge[num].v=y;28     edge[num].w=z;29     edge[num].nxt=head[x];30     head[x]=num++;31 }32 int dis[MAXN];33 int vis[MAXN];34 int happen[MAXN];35 bool SPFA(int bg)36 {37     vis[bg]=1;38     for(int i=head[bg];i!=-1;i=edge[i].nxt)39     {40         if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)41         {42             dis[edge[i].v]=dis[edge[i].u]+edge[i].w;43             if(vis[edge[i].v])44                 return false;45             else if(!SPFA(edge[i].v))46                 return false;47         }48     }49     vis[bg]=0;50     return true;51 }52 int main()53 {54     read(n);read(m);55     for(int i=0;i<=n;i++)56         head[i]=-1;57     for(int i=1;i<=m;i++)58     {59         int how,a,b,c;60         read(how);61         if(how==1)62         {63             read(a);read(b);read(c);64             add_edge(b,a,c);65         }66         else if(how==2)67         {68             read(a);read(b);read(c);69             add_edge(a,b,-c);70         }71         else if(how==3)72         {73             read(a);read(b);74             add_edge(a,b,0);75             add_edge(b,a,0);76         }77     }78     for(int i=0;i<=n;i++)79         dis[i]=maxn;80     dis[0]=0;81     for(int i=1;i<=n;i++)82         add_edge(0,i,1);83     if(SPFA(0))84         printf("Yes");85     else 86         printf("No");87     return 0;88 }