Sol

发现\(NOIP2017\)还没\(AK\)???

赶紧改

考场上明明打出了\(DP\)，没时间了，没判环，重点是没初始化数组，爆\(0\)

\(TAT\)

先最短路，然后\(f[i][j]\)表示到\(i\)时，比最短路大\(j\)的方案

大力记搜就好了

判环就记录一下当前转移的是否在栈中就没了

明明这么简单，可我就是与\(AC\)擦肩而过

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

const int _(1e5 + 5);

typedef long long ll;

typedef int Arr[_];

IL int Input(){

    RG int x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

Arr first, dis, vis, f[55], in[55];

int n, m, k, p, cnt, ans, flg;

struct Edge{

	int to, next, w;

} edge[_ << 1];

queue <int> Q;

IL void Add(RG int u, RG int v, RG int w){

    edge[cnt] = (Edge){v, first[u], w}; first[u] = cnt++;

}

IL void SPFA(){

	dis[1] = 0, Q.push(1), vis[1] = 1;

	while(!Q.empty()){

		RG int u = Q.front(); Q.pop();

		for(RG int e = first[u]; e != -1; e = edge[e].next){

			RG int v = edge[e].to, w = edge[e].w;

			if(dis[u] + w < dis[v]){

				dis[v] = dis[u] + w;

				if(!vis[v]) Q.push(v), vis[v] = 1;

			}

		}

		vis[u] = 0;

	}

}

IL int Dfs(RG int u, RG int d){

	if(flg) return 0;

	if(in[d][u]) return !(flg = 1);

	if(~f[d][u]) return f[d][u];

	in[d][u] = 1, f[d][u] = (u == n && d <= k);

	for(RG int e = first[u]; e != -1; e = edge[e].next){

		RG int v = edge[e].to, fd = dis[u] + edge[e].w + d - dis[v];

		if(fd < 0 || fd > k) continue;

		(f[d][u] += Dfs(v, fd)) %= p;

	}

	return f[d][u] + (in[d][u] = 0);

}

int main(RG int argc, RG char* argv[]){

    RG int T = Input();

    while(T--){

        n = Input(), m = Input(), k = Input(), p = Input();

		Fill(first, -1), Fill(dis, 127), Fill(f, -1), Fill(in, 0);

        flg = ans = cnt = 0;

        for(RG int i = 1; i <= m; i++){

            RG int u = Input(), v = Input(), w = Input();

            Add(u, v, w);

        }

		SPFA(), ans = Dfs(1, 0);

		printf("%d\n", flg ? -1 : ans);

    }

    return 0;

}