J drizzle subway (hierarchical shortest)

Links: https://ac.nowcoder.com/acm/contest/949/J
Source: Cattle-off network

City where a total of rain $\mathrm{m\; subway\; lines,\; respectively\; designated\; Line\; 1,\; Line\; 2,\; ......,\; m\; Line.\; The\; city\; a\; total\; of\; n\; stations,\; numbered\; 1\; ~\; n\; .\; Wherein\; the\; line\; number\; i\; takes\; a\; sitting\; A\; i\; price\; per\; one\; stop\; need\; to\; spend\; more\; bi\; price.\; Line\; i\; have\; C\; i\; stations,\; and\; that\; C\; i\; stations\; are\; known,\; if\; a\; subway\; line\; through\; a\; plurality\; of\; stations,\; it\; can\; take\; the\; subway\; line\; to\; another\; in\; the\; station,\; and\; can\; transfer\; a\; plurality\; of\; times\; .\; Now\; I\; want\; to\; drizzle\; from\; subway\; stations\; to\; the\; s\; t\; stations,\; subway\; waiting\; time\; is\; negligible,\; find\; the\; least\; cost\; price,\; and\; if\; it\; can\; not\; reach\; the\; output\; -1\; .\; (The\; subway\; is\; two-way,\; so\; s\; may\; be\; greater\; than\; t\; )}$

Enter a description:

The first line of the input four positive integers $\mathrm{n-\; ,\; m\; ,\; S\; ,\; T\; respectively\; represent\; the\; number\; of\; stations,\; subway\; line\; number,\; starting\; point\; and\; the\; terminal.\; The\; second\; row\; to\; m\; +\; .\; 1\; line,\; the\; number\; of\; each\; row\; of\; three\; A\; i\; ,\; B\; i\; ,\; C\; i\; ,\; i\; represent\; take\; Line\; price,\; number\; of\; lines\; per\; i\; spend\; one\; stop\; price,\; i\; Line\; The\; number\; of\; stations.\; Next,\; C\; i\; number,\; i\; represents\; the\; number\; of\; each\; station\; line\; numbers\; monotonically.}$

Output Description:

Total row number indicates a minimum cost, if it can not reach the output of -1 .

Example 1

Entry

5 2 1 4
2 2 3 1 3 5
2 1 4 2 3 4 5

Export

7

Explanation

Take the Line 1: In 2; 

$\mathrm{1\; \to \; .\; 3\; :\; In\; 2;\; Transfer\; Line\; 2:\; take\; 2;\; .\; 3\; \to \; .\; 4\; :\; In\; 1;\; therefore\; the\; minimum\; total\; cost\; of\; 7.}$

Remarks:

$. 1 ≤ n- ≤ 10 00 , . 1 ≤ m ≤ 500 , . 1 ≤ S , T ≤ n- . 1 ≤ A I , B I ≤ 100 , . 1 ≤ C I ≤ n- , m [Sigma I = . 1 C I ≤ 10 . 5 ideas: Per underground lines can be seen in FIG layer, between a normal built layer by layer edge to FIG. It may be connected between each layer, how to deal with? Corresponding to each station to establish a virtual point. As long as each station corresponding to the virtual connection point. Is a virtual connection point is connected to the station to spend money on the subway (metro or transfer money). The station is connected to the virtual point takes 0, the effect achieved in transit.Benefits of established virtual point: not according to the attached edge between layers, so each must be connected, it is necessary m * m * n edges, as long as the established virtual point n * m * 2. Calculation is also convenient, but the shortest path from s to find virtual point t is the result of the virtual point. No virtual point, correspond to each station to each subway station s t metro for the most short-circuited. Code:$

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=2100;
struct node{
    int u,v,w,next;
}e[maxn*maxn]; 
int head[maxn*maxn],n,m,s,t,cnt;
int a[maxn],b[maxn],d[maxn*maxn],vis[maxn*maxn];
void add(int u,int v,int w)
{
    e[cnt]={u,v,w,head[u]};
    head[u]=cnt++;
}

void spfa(int st)//spfa
{
    d[st]=0;
    queue<int> q;
    q.push(st);
    vis[st]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].v;
            int w=e[i].w;
            if(d[v]>d[u]+w)
            {
                d[v]=d[u]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        } 
    }        
}

int main()
{
    int num,x,last;
    cnt=0;
    memset(head,-1,sizeof(head));
    memset(d,0x3f,sizeof(d));
    scanf("%d%d%d%d",&n,&m,&s,&t);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&num);
        for(int j=1;j<=num;j++)
        {
            scanf("D% " , & X);
             IF (! J = . 1 ) 
            { 
                the Add ((I - . 1 ) * + n-X, (I- . 1 ) * n-+ Last, B [I]); // same line station is connected to 
                the Add ((I- . 1 ) * n-+ Last, (I- . 1 ) * + n- X, B [I]); 
            } 
            the Add ((I - . 1 ) * + n-X, n-m + X *, 0 ); // associated with each virtual point 
            the Add (n-m * + X, (I- . 1 ) * + n- X, a [I]); 
            Last = X; // previous recording stop 
        } 
    } 
    SPFA (n-*m+s);//最短路 
    if(d[n*m+t]<inf)
        printf("%d\n",d[n*m+t]);
    else
        printf("-1\n");
    return 0;
}