dinic 6个点TLE 求助……

#include <cstdio>
#include <queue>
#include <memory.h>
using namespace std;
const int MAXN = 10000;
const int INF = 1<<29;
int n, m, tot=1, front[MAXN], layer[MAXN]; //1~n:x n+1~2n:y 2n+1:s 2n+2:t
int s, t, ans=0;
bool is_start[MAXN];
struct tEdge
{
    int v, next;
    int c, f;
    inline void addEdge(int tmpu, int tmpv, int tmpc)
    {
        v = tmpv; c = tmpc;
        next = front[tmpu]; 
        front[tmpu] = tot;
        f = 0;
    } 
} e[MAXN*10];
bool bfs()
{
    queue <int> q;
    memset(layer, 0, sizeof(layer));
    layer[s] = 1; q.push(s);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int i=front[u]; i>0; i=e[i].next)
        {
            int v = e[i].v, maxflow = e[i].c - e[i].f; 
            if(layer[v] != 0) continue;
            if(maxflow <= 0) continue;
            layer[v] = layer[u] +1;
            q.push(v);
        }
    }
    if(layer[t] == 0) return false;
    return true;
}
int dfs(int u, int curflow)
{
    if(u == t || curflow == 0) return curflow; 
    int flow = 0;
    for(int i=front[u]; i>0; i=e[i].next)
    {
        int v = e[i].v, maxflow = e[i].c - e[i].f;
        if(layer[v] != layer[u] +1) continue;
        if(maxflow <= 0) continue;
        int addflow = dfs(v, min(curflow, maxflow));
        flow += addflow;
        curflow -= addflow;
        e[i].f += addflow;
        e[i^1].f -= addflow;
        if(curflow == 0) break;
    }
    return flow;
}
int dinic()
{
    int flow = 0;
    while(bfs() == true)  
        flow += dfs(s, INF);
    return flow;
}
void output_dfs(int u)
{
    printf("%d ", u);
    for(int i=front[u]; i>0; i=e[i].next)
    {
        int v = e[i].v;
        if(e[i].f != 1) continue;
        output_dfs(v - n);
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    s = 2*n+1; t = s*n+2;
    for(int i=1; i<=n; i++) //连接原点 汇点 
    {
        e[++tot].addEdge(s, i, 1);
        e[++tot].addEdge(i, s, 0);
        e[++tot].addEdge(n+i, t, 1);
        e[++tot].addEdge(t, n+1, 0);
    }
    for(int i=1; i<=m; i++) //拆点 连边 
    {
        int tmpx, tmpy;
        scanf("%d%d", &tmpx, &tmpy);
        e[++tot].addEdge(tmpx, tmpy+n, 1);
        e[++tot].addEdge(tmpy+n, tmpx, 0);
    }
    int maxflow = dinic(); //dinic求最大流 
    memset(is_start, true, sizeof(is_start));
    for(int i=2; i<=tot; i++) //输出方案 
        if(e[i].f == 1 && e[i].v >n && e[i].v <=2*n)
            is_start[e[i].v - n] = false; //如果一个点不是一条路径的起点，则标记为false 
    for(int i=1; i<=n; i++)
        if(is_start[i] == true) 
            { output_dfs(i); ans ++; printf("\n"); } //从每个路径的节点深搜并输出该路径，每有一个路径ans +1; 
    printf("%d\n", ans); //输出答案 
    return 0;
}