题意：

$N\le 10^5个人，M\le 10^5条拓扑关系$
$问最早第几条边加入的时候可以唯一确定拓扑关系$

分析：

$显然的二分答案啦$
$拓扑排序每次取出一个顶点的时候看看这个顶点是不是唯一的就可以判断拓扑关系是否唯一了$
$时间复杂度O((n+m)logm)$

代码：

//
//  Created by TaoSama on 2016-03-19
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
int u[N], v[N];
vector<int> G[N];

bool check(int m) {
    for(int i = 1; i <= n; ++i) G[i].clear();
    vector<int> in(n + 1, 0);
    for(int i = 1; i <= m; ++i) {
        G[u[i]].push_back(v[i]);
        ++in[v[i]];
    }
    queue<int> q;
    for(int i = 1; i <= n; ++i) if(!in[i]) q.push(i);
    while(q.size()) {
        int u = q.front(); q.pop();
        if(q.size()) return false;
        for(int v : G[u])
            if(--in[v] == 0) q.push(v);
    }
    return true;
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i) scanf("%d%d", u + i, v + i);
    int l = 1, r = m;
    while(l <= r) {
        int m = l + r >> 1;
        if(check(m)) r = m - 1;
        else l = m + 1;
    }
    if(l == m + 1) l = -1;
    printf("%d\n", l);
    return 0;
}