看到求方案数，应该很容易想到dp
f[u][i]表示到点u，且比到u的最短距离多i的方案数
那么需要先预处理dis数组，spfa或者堆优化的dijk
因为考虑到dp的顺序，f[u][i]转移到f[v][j]时，j不可能小于i
所以需要从0到k枚举i，然后从最后一个点开始记忆化搜索
至于判断0环，只需要在记忆化搜索的时候加一个栈即可
1A的代码，哈哈
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#define N 200001 using namespace std; int T, n, m, k, p, cnt, cnt1, ans;
int head[N], to[N], nex[N], val[N], head1[N], to1[N], nex1[N], val1[N], f[N][51], dis[N];
bool flag, vis[N], vis1[N][51], ins[N][51];
queue <int> q; inline int read()
{
int x = 0, f = 1;
char ch = getchar();
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
return x * f;
} inline void add(int x, int y, int z)
{
to[cnt] = y;
val[cnt] = z;
nex[cnt] = head[x];
head[x] = cnt++;
} inline void add1(int x, int y, int z)
{
to1[cnt1] = y;
val1[cnt1] = z;
nex1[cnt1] = head1[x];
head1[x] = cnt1++;
} inline void spfa()
{
int i, v, u;
q.push(1);
dis[1] = 0;
while(!q.empty())
{
u = q.front();
q.pop();
vis[u] = 0;
for(i = head[u]; ~i; i = nex[i])
{
v = to[i];
if(dis[v] > dis[u] + val[i])
{
dis[v] = dis[u] + val[i];
if(!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
} inline void init()
{
int i, x, y, z;
n = read();
m = read();
k = read();
p = read();
for(i = 1; i <= m; i++)
{
x = read();
y = read();
z = read();
add(x, y, z);
add1(y, x, z);
}
spfa();
f[1][0] = 1;
} inline void clear()
{
cnt = cnt1 = ans = flag = 0;
memset(head, -1, sizeof(head));
memset(head1, -1, sizeof(head1));
memset(vis, 0, sizeof(vis));
memset(dis, 127, sizeof(dis));
memset(f, 0, sizeof(f));
memset(vis1, 0, sizeof(vis1));
memset(ins, 0, sizeof(ins));
} inline int dfs(int u, int i)
{
int j, v;
if(i < 0 || flag) return 0;
if(ins[u][i]) return flag = 1;
if(vis1[u][i]) return f[u][i];
ins[u][i] = 1;
for(j = head1[u]; ~j; j = nex1[j])
{
v = to1[j];
f[u][i] = (f[u][i] + dfs(v, dis[u] + i - val1[j] - dis[v])) % p;
}
vis1[u][i] = 1;
ins[u][i] = 0;
return f[u][i];
} inline int solve()
{
int i, t;
for(i = 0; i <= k; i++)
{
t = dfs(n, i);
if(flag) return -1;
ans = (ans + t) % p;
}
return ans;
} int main()
{
T = read();
while(T--)
{
clear();
init();
printf("%d\n", solve());
}
return 0;
}