分析：核心是要計算出任意x天不改變路線所需的花費，即此處的dis[i][j]，再用動態規劃搞一下就可以了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

const int maxn = 105;
const int INF = 1e9;
typedef struct edge
{
	int from, to, dist;
	edge(int u, int v, int d) :from(u), to(v), dist(d) {}
}edge;
typedef struct heapnode
{
	int d, u;
	bool operator <(const heapnode& rhs) const {return d > rhs.d;}
}heapnode;

int days,m,K,e;
int avl[maxn][maxn];
int dis[maxn][maxn];
int can[maxn];
int dp[maxn];

typedef struct Dijkstra
{
	int n, m;
	vector<edge>edges;
	vector<int>G[maxn];
	bool done[maxn];
	int d[maxn];
	int p[maxn];

	void init(int n)
	{
		this->n = n;
		for(int i=0;i<n;i++) G[i].clear();
		edges.clear();
	}

	void add_edge(int from, int to, int dist)
	{
		edges.push_back(edge(from, to, dist));
		m = edges.size();
		G[from].push_back(m - 1);
	}

	void dijkstra(int s)
	{
		priority_queue<heapnode>Q;
		for(int i=0;i<n;i++) d[i] = INF;
		d[s] = 0;
		memset(done,0,sizeof(done));
		Q.push(heapnode{ 0,s });
		while (!Q.empty())
		{
			heapnode x = Q.top();
			Q.pop();
			int u = x.u;
			if (done[u]) continue;
			done[u] = true;
			for(int i=0;i<G[u].size();i++)
			{
				edge &e = edges[G[u][i]];
				if (d[e.to] > d[u] + e.dist&&!can[e.to])
				{
					d[e.to] = d[u] + e.dist;
					p[e.to] = G[u][i];
					Q.push(heapnode{ d[e.to],e.to });
				}
			}
		}
	}
}Dijkstra;

Dijkstra D;
Dijkstra *f=&D;
void solve()
{
    for(int i=0; i<maxn; i++)
        for(int j=0; j<maxn; j++)
            dis[i][j]=INF;
    for(int i=1; i<=days; i++)
    {
        for(int j=i; j<=days; j++)
        {
            memset(can,0,sizeof(can));
            for(int k=0; k<m; k++)
                for(int g=i; g<=j; g++)
                    can[k]|=avl[k][g];
            f->dijkstra(0);
            dis[i][j]=f->d[m-1];
            if(dis[i][j]<INF)
                dis[i][j]*=(j-i+1);
        }
    }
}
void DP()
{
    for(int i=1;i<=days;i++)
        dp[i]=dis[1][i];
    for(int i=2;i<=days;i++)
        for(int j=1;j<i;j++)
            dp[i]=min(dp[i],dp[j]+dis[j+1][i]+K);
}
int main()
{
	int u, v, w;
	scanf("%d%d%d%d",&days,&m,&K,&e);
	f->init(m);
	for(int i=0;i<e;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		u--,v--;
		f->add_edge(u,v,w);
		f->add_edge(v,u,w);
	}
	memset(avl,0,sizeof(avl));
	int num,st,ed;
	scanf("%d",&num);
	for(int i=0;i<num;i++)
	{
	    scanf("%d%d%d",&u,&st,&ed);
	    u--;
	    for(int j=st;j<=ed;j++)
            avl[u][j]=1;
	}
	solve();
	DP();
	printf("%d\n",dp[days]);
	return 0;
}