NOI Day1T1归程(Kruskal重构树+Dijkstra)

NOI Day1T1归程(Kruskal重构树+Dijkstra)

题目

洛谷题目传送门

题解

其实我不想写......，所以......

挖个坑......我以后一定会补的

luogu的题解讲的还是很详细的......

恩，感谢cwen详细教我做这道题，让我2遍A

然后我丑陋的代码(代码太长，所以压得很丑)

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<ctime>
#include<queue>
#include<stack>
#include<vector>
#define rg register
#define il inline
#define lst long long
#define ldb long double
#define N 400050
#define M 800050
using namespace std;
const lst Inf=1e16;
il int read()
{
    rg int s=0,m=0;rg char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')m=1;ch=getchar();}
    while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
    return m?-s:s;
}

int n,m,Q,K,S,u,p;
int cnt,tot;
lst ans;
int hd[N],fa[N];
lst dis[N],tr_dis[N];
lst f[N][25],g[N][25];
struct WAY{int u,v;lst w,h;}edge[M];
struct EDGE{int to,nxt;lst v,h;}ljl[M<<1];
il int operator<(rg const WAY &a,rg const WAY &b){return a.h>b.h;}

il void memt()
{
    cnt=0;
    memset(hd,0,sizeof(hd));
    memset(ljl,0,sizeof(ljl));
}
il void add(rg int p,rg int q,rg int o,rg int hi){ljl[++cnt]=(EDGE){q,hd[p],o,hi},hd[p]=cnt;}
int vis[N];
struct hh{lst D;int id;};
bool operator<(rg const hh &a,rg const hh &b){return a.D>b.D;}
priority_queue <hh> H;
il void Dijkstra()
{
    for(rg int i=1;i<=n;++i)vis[i]=0,dis[i]=Inf;
    dis[1]=0,H.push((hh){0,1});
    while(!H.empty())
    {
        rg hh S=H.top();H.pop();
        rg int now=S.id;
        if(vis[now])continue;
        vis[now]=1;
        for(rg int i=hd[now];i;i=ljl[i].nxt)
        {
            rg int qw=ljl[i].to;
            if(dis[qw]>dis[now]+ljl[i].v)
            {
                dis[qw]=dis[now]+ljl[i].v;
                H.push((hh){dis[qw],qw});
            }
        }
    }
}
int find_fa(rg int now)
{
    if(fa[now]==now)return now;
    return fa[now]=find_fa(fa[now]);
}
il void Kruskal()
{
    rg int nn=2*n;tot=n;
    for(rg int i=1;i<=nn;++i)fa[i]=i;
    sort(edge+1,edge+m+1);
    for(rg int i=1;i<=m;++i)
    {
        rg int u=edge[i].u,v=edge[i].v;
        rg int fx=find_fa(u),fy=find_fa(v);
        if(fx!=fy)
        {
            tot++;
            add(tot,fx,0,edge[i].h);
            add(tot,fy,0,edge[i].h);
            fa[fx]=fa[fy]=tot;
            if(tot==nn-1)break;
        }
    }
}

void dfs(rg int now,rg int fm,rg lst up)
{
    f[now][0]=fm,g[now][0]=up;
    for(rg int j=1;j<=20;++j)
    {
        f[now][j]=f[f[now][j-1]][j-1];
        g[now][j]=min(g[now][j-1],g[f[now][j-1]][j-1]);
    }
    for(rg int i=hd[now];i;i=ljl[i].nxt)
    {
        rg int qw=ljl[i].to;
        if(qw!=fm)
            dfs(qw,now,ljl[i].h);
        tr_dis[now]=min(tr_dis[qw],tr_dis[now]);
    }
}

int main()
{
    for(rg int T=read();T;T--)
    {
        memt();
        n=read(),m=read();
        for(rg int i=1;i<=m;++i)
        {
            rg int p=read(),q=read(),o=read(),hi=read();
            add(p,q,o,hi),add(q,p,o,hi);
            edge[i]=(WAY){p,q,o,hi};
        }
        Dijkstra(),memt();
        Kruskal();
        for(rg int i=1;i<=n;++i)tr_dis[i]=dis[i];
        for(rg int i=n+1;i<=tot;++i)tr_dis[i]=Inf;
        dfs(tot,0,0);
        ans=0;
        Q=read(),K=read(),S=read();
        while(Q--)
        {
            u=read(),p=read();
            u=(u+K*ans-1)%n+1;
            p=(p+K*ans)%(S+1);
            for(rg int j=19;j>=0;--j)
                if(g[u][j]>p)u=f[u][j];
            ans=tr_dis[u];
            printf("%lld\n",ans);
        }
    }
    return 0;
}