这题很容易想到一个做法，缩点以后去掉重边，看看是不是一条链。
但是我这样打以后TLE了，因为边去重太慢了，所以我就改成拓扑，如果一次入队的点数超过了1，就不满足条件。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;

const int N = 1010, M = 6010;

int n, m;
int tot, Head[N], ver[M], Next[M];
int tp, sta[N], id, dfn[N], low[N], cnt, dcc[N]; 
int in[N]; bool ins[N];
vector<int> a[N];

void add(int x, int y) {
    tot++;
    ver[tot] = y;
    Next[tot] = Head[x];
    Head[x] = tot;
}

void v_add(int x, int y) {
    a[x].push_back(y);
}

void Tarjan(int x) {
    sta[++tp] = x;
    dfn[x] = low[x] = ++id;
    ins[x] = 1;
    for (int i = Head[x]; i; i = Next[i]) {
        int y = ver[i];
        if (!dfn[y]) Tarjan(y), low[x] = min(low[x], low[y]);
        else if (ins[y]) low[x] = min(low[x], dfn[y]);
    }
    if (low[x] == dfn[x]) {
        cnt++; int y;
        do {
            y = sta[tp--];
            dcc[y] = cnt;
            ins[y] = 0;
        } while (x != y);
    }
}

int main() {
    int T; cin >> T;
    while (T--) {
        tot = 0, memset(Head, 0, sizeof(Head));
        id = 0, memset(dfn, 0, sizeof(dfn));
        memset(low, 0, sizeof(low));
        for (int i = 1; i <= cnt; i++) a[i].clear();
        cnt = 0; memset(dcc, 0, sizeof(dcc));
        memset(in, 0, sizeof(in));
        cin >> n >> m;
        for (int i = 1; i <= m; i++) {
            int x, y; 
            scanf("%d%d", &x, &y);
            add(x, y); 
        }
        for (int i = 1; i <= n; i++)
            if (!dfn[i]) Tarjan(i);
        for (int x = 1; x <= n; x++)
            for (int i = Head[x]; i; i = Next[i]) {
                int y = ver[i];
                if (dcc[x] == dcc[y]) continue;
                in[dcc[y]]++; v_add(dcc[x], dcc[y]);
            }
        queue<int> q;
        for (int i = 1; i <= cnt; i++)
            if (!in[i]) q.push(i);
        bool bk = 1;
        while (q.size()) {
            int x = q.front(); q.pop();
            if (q.size() > 0) { bk = 0; break; }
            for (int i = 0; i < a[x].size(); i++) {
                int y = a[x][i];
                if (!--in[y]) q.push(y);
            }
        }
        puts(bk ? "Yes" : "No");
    }
    return 0;
}