Luogu4768 NOI2018 归程 最短路、Kruskal重构树

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题意：给出一个$N$个点、$M$条边的图，每条边有长度和海拔，$Q$组询问，每一次询问从$v$开始，经过海拔超过$p$的边所能到达的所有点中到点$1$的最短路的最小值，强制在线。$N \leq 2 \times 10^5 , M , Q \leq 4 \times 10^5$

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关于$SPFA...$

与边的权值有关的连通块问题，经常可以考虑到$Kruskal$重构树。我们以海拔从大到小构建$Kruskal$重构树，那么对于每一次询问，可以到达的点就对应$Kruskal$重构树上的一棵子树。我们对于每一个点记录它的子树的叶子节点的最短路的最小值，每一次倍增找到询问对应的那一棵子树就能得到答案了。

#include<bits/stdc++.h>
//This code is written by Itst
using namespace std;

inline int read(){
    ;
    ;
    char c = getchar();
    while(c != EOF && !isdigit(c)){
        if(c == '-')
            f = ;
        c = getchar();
    }
    while(c != EOF && isdigit(c)){
        a = (a << ) + (a << ) + (c ^ ');
        c = getchar();
    }
    return f ? -a : a;
}

;
struct edge{
    int start , end , hei;
}now[MAXN];
struct Edge{
    int end , upEd , w;
}Ed[MAXN << ];
] , jump[MAXN << ][] , val[MAXN << ] , ans[MAXN << ] , ch[MAXN << ][] , cntNode , cntEd , N , M , Q;
priority_queue < pair < int , int > > q;

bool operator <(edge a , edge b){
    return a.hei > b.hei;
}

int find(int a){
    return fa[a] == a ? a : (fa[a] = find(fa[a]));
}

inline void addEd(int a , int b , int c){
    Ed[++cntEd].end = b;
    Ed[cntEd].upEd = head[a];
    Ed[cntEd].w = c;
    head[a] = cntEd;
}

void Dijk(){
    memset(minDis , 0x7f , sizeof(minDis));
    minDis[] = ;
    q.push(make_pair( , ));
    while(!q.empty()){
        pair < int , int > t = q.top();
        q.pop();
        if(-t.first > minDis[t.second])
            continue;
        for(int i = head[t.second] ; i ; i = Ed[i].upEd)
            if(minDis[Ed[i].end] > minDis[t.second] + Ed[i].w){
                minDis[Ed[i].end] = minDis[t.second] + Ed[i].w;
                q.push(make_pair(-minDis[Ed[i].end] , Ed[i].end));
            }
    }
}

void dfs(int node){
    if(!node)
        return;
     ; i <=  && jump[node][i - ] ; ++i)
        jump[node][i] = jump[jump[node][i - ]][i - ];
    dfs(ch[node][]);
    dfs(ch[node][]);
}

inline int jumpToAll(int x , int h){
     ; i >=  ; --i)
        if(val[jump[x][i]] > h)
            x = jump[x][i];
    return x;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("4768.in" , "r" , stdin);
    //freopen("4768.out" , "w" , stdout);
#endif
    for(int T = read() ; T ; --T){
        memset(jump ,  , sizeof(jump));
        memset(ch ,  , sizeof(ch));
        memset(head ,  , sizeof(head));
        cntEd = ;
        N = read();
        M = read();
         ; i <= M ; i++){
            int a = read() , b = read() , c = read() , d = read();
            now[i].start = a;
            now[i].end = b;
            now[i].hei = d;
            addEd(a , b , c);
            addEd(b , a , c);
        }
        Dijk();
         ; i <= N ; i++){
            fa[i] = i;
            ans[i] = minDis[i];
        }
        sort(now +  , now + M + );
        cntNode = N;
         ; i <= M ; ++i)
            if(find(now[i].start) != find(now[i].end)){
                int a = find(now[i].start) , b = find(now[i].end);
                fa[a] = fa[b] = jump[a][] = jump[b][] = ++cntNode;
                ch[cntNode][] = a;
                ch[cntNode][] = b;
                val[cntNode] = now[i].hei;
                ans[cntNode] = min(ans[a] , ans[b]);
                fa[cntNode] = cntNode;
            }
        dfs(cntNode);
         , Q = read() , K = read() , S = read();
        while(Q--){
            int a = read() , b = read();
            a = (0ll + a + K * lastans - ) % N + ;
            b = (0ll + b + K * lastans) % (S + );
            int t = jumpToAll(a , b);
            printf("%d\n" , lastans = ans[t]);
        }
    }
    ;
}