【BZOJ5415&UOJ393】归程（Kruskal重构树，最短路）

题意：From https://www.cnblogs.com/Memory-of-winter/p/11628351.html

思路：先从1开始跑一遍dijkstra，建出kruskal重构树之后每个叶子结点的权值为它到1的距离

询问等价于从v开始只要倍增的点的权值>p就往上跳，这样跳到某个点u之后询问u的子树中叶子结点最小的权值

因为是静态的，实际上可以不把kruskal实际建出来，只要维护倍增数组和子树中最小值即可

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef long double ld;
 typedef pair<int,int> PII;
 typedef pair<ll,int> Pli;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,ll>P;
 #define N  400000+10
 #define M  800000+10
 #define INF 4e9+10
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
 #define fors(i) for(auto i:e[x]) if(i!=p)

 const int MOD=1e9+,inv2=(MOD+)/;
       //int p=1e6+3;
       //double eps=1e-6;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct edge
 {
     int x,y,l,a;
 }a[M];

 bool cmp(edge a,edge b)
 {
     return a.a>b.a;
 }

 ll dis[N],val[N];
 int f[N][],head[N],vet[M],nxt[M],len[M],vis[N],fa[N],n,m,num,tot;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 ll readll()
 {
    ll v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void add(int a,int b,int c)
 {
     nxt[++tot]=head[a];
     vet[tot]=b;
     len[tot]=c;
     head[a]=tot;
 }

 void dijkstra()
 {
     priority_queue<Pli> q;
     rep(i,,n*) dis[i]=INF,vis[i]=;
     q.push(MP(,)); dis[]=;
     while(!q.empty())
     {
         int u=q.top().se;
         q.pop();
         if(vis[u]) continue;
         vis[u]=;
         int e=head[u];
         while(e)
         {
             int v=vet[e];
             if(dis[u]+len[e]<dis[v])
             {
                 dis[v]=dis[u]+len[e];
                 q.push(MP(-dis[v],v));
             }
             e=nxt[e];
         }
     }
 }

 int dsu(int k)
 {
     if(fa[k]!=k) fa[k]=dsu(fa[k]);
     return fa[k];
 }

 void Add(int x,int y)
 {
     f[y][]=x;
     dis[x]=min(dis[x],dis[y]);
 }

 int calc(int u,ll p)
 {
     per(i,,)
      if(val[f[u][i]]>p) u=f[u][i];
     return u;
 }

 void build()
 {
     num=n;
     rep(i,,*n) fa[i]=i;
     sort(a+,a+m+,cmp);
     rep(i,,m)
     {
         int p=dsu(a[i].x),q=dsu(a[i].y);
         if(p!=q)
         {
             num++;
             fa[p]=fa[q]=num;
             val[num]=a[i].a;
             Add(num,p);
             Add(num,q);
             if(num==*n-) break;
         }
     }
     rep(i,,)
      rep(j,,num) f[j][i]=f[f[j][i-]][i-];
 }

 void solve()
 {
     int q=read(),k=read(),s=read();
     ll lastans=;
     while(q--)
     {
         int v=read();
         ll p=readll();
         if(k==)
         {
             v=(v-)%n+;
             p=p%(s+);
         }
          else
          {
              v=(v+lastans-)%n+;
              p=(p+lastans)%(s+);
          }
         int t=calc(v,p);
         lastans=dis[t];
         printf("%lld\n",lastans);
     }
 }
 int main()
 {
     int cas=read();
     while(cas--)
     {
         n=read(),m=read();
         tot=;
         rep(i,,n) head[i]=;
         rep(i,,m)
         {
             a[i].x=read(),a[i].y=read(),a[i].l=read(),a[i].a=read();
             add(a[i].x,a[i].y,a[i].l);
             add(a[i].y,a[i].x,a[i].l);
         }
         dijkstra();
         build();
         solve();
     }
     return ;
 }