题目大意:给定一张图,对于每条边,询问 1:是否存在于某个最小割集中,2:是否存在于所有最小割集中
对残余网络做一次 tarjan
对于一条满流边,若 id[s] != id[t],则能存在于某个最小割集中
若又有 id[s] == id[S] && id[t] == id[T] 则一定能存在于最小割集中
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define INF 1000000000
#define N 4010
#define M 120500
using namespace std;
int n,m,siz = 1,S,T,tot,cnt,top;
int d[N],p[N],dfn[N],low[N],st[N],id[N];
int first[N],next[M],to[M],len[M];
bool v[N];
void inser(int x,int y,int w)
{
next[++ siz] = first[x];
first[x] = siz;
to[siz] = y;
len[siz] = w;
}
void add_edge(int x,int y,int w)
{
inser(x,y,w),inser(y,x,0);
}
bool bfs()
{
int head = 0,tail = 1;
memset(d,0,sizeof(d));
d[p[1] = S] = 1;
while (head ^ tail)
{
int x = p[++ head];
for (int i = first[x];i;i = next[i])
if (!d[to[i]] && len[i]) d[p[++ tail] = to[i]] = d[x] + 1;
}
return d[T];
}
int dfs(int x,int flow)
{
if (x == T) return flow;
int ret = 0;
for (int i = first[x];i && flow;i = next[i])
if (d[to[i]] == d[x] + 1 && len[i])
{
int w = dfs(to[i],min(flow,len[i]));
len[i] -= w,len[i ^ 1] += w;
flow -= w,ret += w;
}
if (!ret) d[x] = 0;
return ret;
}
int dinic()
{
int ret = 0;
while (bfs()) ret += dfs(S,INF);
return ret;
}
void dfs(int x)
{
dfn[x] = low[x] = ++ cnt;
st[++ top] = x;
v[x] = true;
for (int i = first[x];i;i = next[i])
if (len[i])
{
if (!dfn[to[i]]) dfs(to[i]),low[x] = min(low[x],low[to[i]]);
else if (v[to[i]]) low[x] = min(low[x],low[to[i]]);
}
if (dfn[x] == low[x])
{
tot ++;
while (st[top] ^ x)
id[st[top]] = tot,v[st[top --]] = false;
id[st[top]] = tot,v[st[top --]] = false;
}
}
int main()
{
scanf("%d%d%d%d",&n,&m,&S,&T);
for (int x,y,w,i = 1;i <= m;i ++)
scanf("%d%d%d",&x,&y,&w),add_edge(x,y,w);
dinic();
for (int i = 1;i <= n;i ++)
if (!dfn[i]) dfs(i);
for (int i = 2;i <= siz;i += 2)
{
int x = to[i ^ 1],y = to[i];
printf("%d ",(id[x] != id[y] && !len[i]) ? 1 : 0);
printf("%d\n",(id[x] != id[y] && !len[i] && id[x] == id[S] && id[y] == id[T]) ? 1 : 0);
}
return 0;
}