题目意思

大水题，给出n个顶点的m条边的有向带权图。
求从1到n的方法数有多少条？是不是很震惊，边权不同控制发出时间即可同一时间到达，所以边权的信息没什么用。

解题思路

我写的dfs，回溯法先求道各个入点的路径数，累加就是这个出度点的答案。
使用记忆化搜索可以简化计算时间

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
const int MOD = 20010905;
int head[N], n, m, tot;
struct Node {
    int v, next;
}edge[N << 1];

void add(int u, int v) {
    ++tot;
    edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot;
}

int dp[N];
int dfs(int x) {
    if(dp[x])    return dp[x];
    for (int i = head[x]; i; i = edge[i].next)
        (dp[x] += dfs(edge[i].v)) %= MOD;
    return dp[x];
}

int main() {
    n = read(), m = read();
    for (int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        add(u, v);
    }
    dp[n] = 1;
    print(dfs(1));
    return 0;
}