【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+7,inv2=(MOD+1)/2;
      //int p=1e6+3;
      //double eps=1e-6;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

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][21],head[N],vet[M],nxt[M],len[M],vis[N],fa[N],n,m,num,tot;

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

ll readll()
{
   ll v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,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,1,n*2) dis[i]=INF,vis[i]=0;
    q.push(MP(0,1)); dis[1]=0;
    while(!q.empty())
    {
        int u=q.top().se;
        q.pop();
        if(vis[u]) continue;
        vis[u]=1;
        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][0]=x;
    dis[x]=min(dis[x],dis[y]);
}

int calc(int u,ll p)
{
    per(i,20,0)
     if(val[f[u][i]]>p) u=f[u][i];
    return u;
}

void build()
{
    num=n;
    rep(i,1,2*n) fa[i]=i;
    sort(a+1,a+m+1,cmp);
    rep(i,1,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==2*n-1) break;
        }
    }
    rep(i,1,20)
     rep(j,1,num) f[j][i]=f[f[j][i-1]][i-1];
}

void solve()
{
    int q=read(),k=read(),s=read();
    ll lastans=0;
    while(q--)
    {
        int v=read();
        ll p=readll();
        if(k==0)
        {
            v=(v-1)%n+1;
            p=p%(s+1);
        }
         else
         {
             v=(v+lastans-1)%n+1;
             p=(p+lastans)%(s+1);
         }
        int t=calc(v,p);
        lastans=dis[t];
        printf("%lld\n",lastans);
    }
}
int main()
{
    int cas=read();
    while(cas--)
    {
        n=read(),m=read();
        tot=0;
        rep(i,1,n) head[i]=0;
        rep(i,1,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 0;
}