首先一定跑一遍最小割,在殘量網絡上tarjan一遍
(1) 滿足u和v不在同一個強連通分量且有流通過。
(2)滿足(1)且u在s的強連通分量,v在t的強連通分量。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
stack<int> s1;
queue<int> q;
int read()
{
char ch=getchar();int f=;
while(ch>'9'||ch<'0') ch=getchar();
while(ch>='0'&&ch<='9') {f=f*+(ch^);ch=getchar();}
return f;
}
int n,m,s,t,dis[],head[],dfn[],low[],pos[],size[];
int tim,tot,cnt;long long ans;bool vis[];
struct node
{
int from;
int to;
int next;
int w;
}edge[];
void add(int u,int v,int w)
{
edge[tot].from=u;
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
bool bfs()
{
memset(dis,,sizeof(dis));
dis[s]=;
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
for(int i=head[x];i!=-;i=edge[i].next)
{
if(!dis[edge[i].to]&&edge[i].w)
{
dis[edge[i].to]=dis[x]+;
q.push(edge[i].to);
}
}
}
if(!dis[t])
return ;
return ;
}
int dfs(int x,int flow)
{
if(x==t||!flow)
return flow;
int ret=;
for(int i=head[x];i!=-;i=edge[i].next)
{
if(dis[edge[i].to]==dis[x]+)
{
int f=dfs(edge[i].to,min(flow,edge[i].w));
edge[i].w-=f;edge[i^].w+=f;
ret+=f;
flow-=f;
if(!flow)
break;
}
}
if(!ret)
dis[x]=-;
return ret;
}
void dinic()
{
while(bfs())
{
ans+=dfs(s,);
}
}
void tarjan(int x)
{
dfn[x]=low[x]=++tim;
s1.push(x);
vis[x]=;
for(int i=head[x];i!=-;i=edge[i].next)
{
if(edge[i^].w)
{
if(!dfn[edge[i].to])
{
tarjan(edge[i].to);
low[x]=min(low[x],low[edge[i].to]);
}
else if(vis[edge[i].to])
{
low[x]=min(low[x],dfn[edge[i].to]);
}
}
}
if(low[x]==dfn[x])
{
cnt++;
while(s1.top()!=x)
{
pos[s1.top()]=cnt;
size[cnt]++;
vis[s1.top()]=;
s1.pop();
}
s1.pop();
pos[x]=cnt;
size[cnt]++;
vis[x]=;
}
}
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
memset(head,-,sizeof(head));
n=read(),m=read(),s=read(),t=read();
int x,y,z;
for(int i=;i<=m;i++)
{
x=read(),y=read(),z=read();
add(x,y,z);
add(y,x,);
}
dinic();
for(int i=;i<=n;i++)
{
if(!dfn[i])
tarjan(i);
}
for(int i=;i<tot;i+=)
{
if(pos[edge[i].from]!=pos[edge[i].to]&&!edge[i].w)
{
printf("1 ");
if(pos[s]==pos[edge[i].from]&&pos[t]==pos[edge[i].to])
{
puts("1");
}
else puts("0");
}
else
{
puts("0 0");
}
}
//for(int i=1;i<=n;i++)
//printf("%d ",pos[i]);
}