洛谷$P4768 [NOI2018]$归程 $kruscal$重构树

正解:$kruscal$重构树

解题报告:

传送门$QwQ$

语文不好选手没有人权$TT$连题目都看不懂真的要哭了$kk$

所以先放个题目大意?就说给定一个$n$个点,$m$条边的图,每条边有长度和海拔.有$Q$组询问,每次查询从$x$出发，经过海拔超过$p$的所有路径,能到达的节点中距离1号节点的最短路径长是多少$QwQ$

首先看到这个对海拔的限制就显然考虑$kruscal$重构树呗$QwQ$,然后说是所有海拔超过$p$的路径能到达的点中最短路最小的点$QwQ$?

可以理解成把最短路作为一个点的属性,然后问$kruscal$重构树中一棵子树中的属性$min$是多少$QwQ$?

那不给每个节点记录下子树内的属性$min$然后每次查询就跳下就成$QwQ$?

然后就做完了鸭$QwQ$

$overrrrrr$

然后$attention$,那个$lastans$在每一轮都要清零,,,因为它是说每个第一天$lastans$都等于0$QwQ$

$over$

 

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define mp make_pair
#define int long long
#define P pair<int,int>
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define fy(i) edge_krus[i].wei
#define ri register int
#define rc register char
#define rb register bool
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)

const int N=1000000+1000,M=1000000+10;
int ed_cnt,ed_tr_cnt,head[N],as[N<<1],fat[N<<1][20],nod_cnt,n,m,fa[N<<1],lstas,val[N<<1];
bool vis[N];
struct ed{int to,nxt,wei;}edge[M<<1];
struct ed_tr{int fr,to,wei;}edge_krus[M];

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il bool cmp(ed_tr gd,ed_tr gs){return gd.wei>gs.wei;}
int fd(ri x){return fa[x]==x?x:fa[x]=fd(fa[x]);}
il void ad(ri x,ri y,ri z){edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;}
il void dij()
{
    priority_queue< P,vector<P>,greater<P> >Q;
    memset(vis,0,sizeof(vis));memset(as,63,sizeof(as));as[1]=0;Q.push(mp(0,1));
    while(!Q.empty())
    {
        ri nw=Q.top().second;Q.pop();if(vis[nw])continue;vis[nw]=1;
        e(i,nw)if(as[t(i)]>as[nw]+w(i))as[t(i)]=as[nw]+w(i),Q.push(mp(as[t(i)],t(i)));
    }
}
il int jump(ri x,ri y){my(i,19,0)if(val[fat[x][i]]>y)x=fat[x][i];return x;}

signed main()
{
    freopen("return.in","r",stdin);freopen("return.out","w",stdout);
    ri T=read();
    while(T--)
    {
        n=read();m=read();rp(i,1,n<<1)fa[i]=i;nod_cnt=n;ed_cnt=0;memset(head,0,sizeof(head));lstas=0;
        rp(i,1,m){ri x=read(),y=read(),z=read(),p=read();ad(x,y,z);ad(y,x,z);edge_krus[i]=(ed_tr){x,y,p};}
        dij();sort(edge_krus+1,edge_krus+1+m,cmp);
        //rp(i,1,n)printf("as[%d]=%d
",i,as[i]);
        rp(i,1,m)
        {
            ri x=fd(edge_krus[i].fr),y=fd(edge_krus[i].to);if(x==y)continue;
            //printf("x=%d y=%d z=%d fy=%d
",edge_krus[i].fr,edge_krus[i].to,edge_krus[i].wei,fy(i));
            fa[x]=fa[y]=fat[x][0]=fat[y][0]=++nod_cnt;as[nod_cnt]=min(as[x],as[y]);val[nod_cnt]=fy(i);
            //printf("val[fa[%d]=fa[%d]=%d]=%d nod_cnt=%d
",x,y,fa[x],val[fa[x]],nod_cnt);
        }
        rp(i,1,19)rp(j,1,nod_cnt)fat[j][i]=fat[fat[j][i-1]][i-1];
        ri Q=read(),k=read(),s=read();
        while(Q--)
        {
            ri x=(read()+k*lstas-1)%n+1,y=(read()+k*lstas)%(s+1);printf("%lld
",lstas=as[jump(x,y)]);
        }
    }
    return 0;
}