P3953 [NOIP2017 提高组] 逛公园

// https://www.luogu.com.cn/problem/P3953
/*
dis(1,n)+k>=dist(1,n) 的方案数
30-最短路计数 <- dp
70-没有零环 自然的设 f[u][k] dist(1,u)==dis(1,u)+k 方案数 
dis(1,u)+k+edge(u,v)==dis(1,v)+x x=dis(1,u)+k+edge(u,v)-dis(1,v) k=dis(1,v)+x-dis(1,u)-edge(u,v)
f[v][x]=sum(f[u][ k=dis(1,v)+x-dis(1,u)-edge(u,v)] ) 
由于 正向图不一定能走到 n点 所以建反图 
f[u][k] dist(u,n)==dis(u,n)+k 方案数 即n-u的合法方案数 
dis(u,n)+k-edge(u,v)==dis(n,v)+x   k=dis(n,v)+x-dis(n,u)+edge(u,v) x=dis(u,n)+k-edge(u,v)-dis(n,v)
f[u][k]=sum(f[v][x=dis(u,n)+k-edge(u,v)-dis(n,v)])  f[n][0]=1

100: 有零环 有无数条路径 检查是否已经入栈 

*/
/*
2
5 7 2 10
1 2 1
2 4 0
4 5 2
2 3 2
3 4 1
3 5 2
1 5 3
2 2 0 10
1 2 0
2 1 0

3
-1
*/
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string.h>
//#include<queue>
//#include<vector>
#include<bits/stdc++.h>
#define ll long long
#define ddd printf("-----------------------\n");
using namespace std;
const int maxn=2e5 +10;
const int mod=998244353;
const int inf=0x3f3f3f3f;

int n,m,k,p;
int head1[maxn],tot1,tot2,head2[maxn],to1[maxn<<1],to2[maxn<<1],nxt1[maxn<<1],nxt2[maxn<<1],w1[maxn<<1],w2[maxn<<1];       
int f[maxn>>1][55],ins[maxn>>1][60],dis[maxn],vis[maxn],ans,flag;

void add1(int a,int b,int val){
    to1[++tot1]=b,nxt1[tot1]=head1[a],head1[a]=tot1,w1[tot1]=val;
}
void add2(int a,int b,int val){
    to2[++tot2]=b,nxt2[tot2]=head2[a],head2[a]=tot2,w2[tot2]=val;
}

void spfa()
{
    memset(vis,0,sizeof(vis));
    memset(dis,inf,sizeof(dis));
    queue<int> q;
    q.push(n),dis[n]=0,vis[n]=1;
    
    while(q.size())
    {
        int u=q.front(); q.pop(); vis[u]=0;
        for(int i=head2[u];i;i=nxt2[i])
        {
            int v=to2[i];
            if(dis[v]>dis[u]+w2[i])
            {
                dis[v]=dis[u]+w1[i];
                if(vis[v]==0) q.push(v),vis[v]=1;
            }
        }
    }
}

int dfs(int u,int know)
{
    if(ins[u][know]){ flag=1; return 0;}//!!!!!!!!!!!!!!!!
    if(f[u][know]) return f[u][know];
    ins[u][know]=1;
    
    int sum=0;
    for(int i=head1[u];i;i=nxt1[i])
    {
        int v=to1[i];
        int tmp=know-(w1[i]+dis[v]-dis[u]);
        if(tmp<0||tmp>k) continue;//!!!!!!!!!!!!!!!!!!!!!!
        sum=(sum+dfs(v,tmp))%p;
        if(flag) return 0;
    }
    
    if(u==n&&know==0) sum=1;
    ins[u][know]=0;
    return f[u][know]=sum;
}

int main()
{
    ios::sync_with_stdio(false);
    int T;cin>>T;
    while(T--)
    {
        memset(head1,0,sizeof(head1)),memset(head2,0,sizeof(head2)); tot1=0,tot2=0;
        memset(to1,0,sizeof(to1)),memset(to2,0,sizeof(to2));
        memset(w1,0,sizeof(w1)),memset(w2,0,sizeof(w2));
        memset(nxt1,0,sizeof(nxt1)),memset(nxt2,0,sizeof(nxt2));
        memset(ins,0,sizeof(ins)),memset(f,0,sizeof(f));
        flag=0,ans=0;
        
        cin>>n>>m>>k>>p;
        for(int i=1;i<=m;i++){
            int a,b,c;cin>>a>>b>>c;
            add1(a,b,c),add2(b,a,c);
        }
        
        spfa();
        
        for(int i=0;i<=k;i++) ans=(ans+dfs(1,i))%p;
        if(flag==1) cout<<"-1\n";
        else cout<<ans<<'\n';
        
    } 
    return 0;
}