[NOI2018]归程 [Kruskal 重构树]

传送门

刚刚学Kruskal重构树就来写这道题, 我都佩服我自己...

不过还好把Kruskal 重构树学会了

https://blog.csdn.net/niiick/article/details/81952126

关于本题, 题意: 将v到1的路径分成两半, v-v的路u的海拔最小的至少为a+1, 求u到1的最小值

以下来自https://blog.csdn.net/niiick/article/details/81138433 

#include<bits/stdc++.h>
#define N 400050
#define M N*2
#define LL long long
#define inf 1000000000000000
using namespace std;

int first[N],next[M],to[M],w[M],tot;  // Kruskal 重构树的边 & dij

struct Node{int u,v,w;}E[M]; int val[N],f[N][23]; LL Min_dis[N];// Kruskal 重构树
bool cmp(Node a,Node b){return a.w>b.w;}

LL dis[N]; // dijsktra
int T,n,m,q,k,s,cnt; LL ans;
int fa[N]; int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}

int read(){
	int cnt=0; char ch=0;
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))cnt=cnt*10+(ch-'0'),ch=getchar();
	return cnt;
}

void add(int x,int y,int z){next[++tot]=first[x],first[x]=tot,to[tot]=y,w[tot]=z;}
void Init_graph(){
	memset(first,0,sizeof(first));
	memset(next,0,sizeof(next));
	memset(to,0,sizeof(to));
	memset(w,0,sizeof(w)); tot = 0;
}
void dijsktra(){
	priority_queue<pair<int,int> >q;
	for(int i=1;i<=n;i++) dis[i] = inf;
	dis[1]=0; q.push(make_pair(0,1));
	while(!q.empty()){
		int x=q.top().second; q.pop();
		for(int i=first[x];i;i=next[i]){
			int t=to[i]; 
			if(dis[t] > dis[x] + w[i]){
				dis[t] = dis[x] + w[i];
				q.push(make_pair(-dis[t],t));
			}
		}
	} 
}

void Kruskal(){
	Init_graph();
	sort(E+1,E+m+1,cmp);
	cnt = n;
	for(int i=1;i<=m;i++){
		int x = E[i].u, y = E[i].v;
		int fx = find(x), fy = find(y);
		if(fx!=fy){
			val[++cnt] = E[i].w;
			fa[fx] = fa[fy] = cnt;
			add(cnt,fx,0); add(cnt,fy,0);
		}
	}
}

void dfs(int u){
	if(u<=n) Min_dis[u] = dis[u];
	for(int i=1;i<=20;i++) 
		f[u][i] = f[f[u][i-1]][i-1];
	for(int i=first[u];i;i=next[i]){
		int t=to[i]; f[t][0] = u;
		dfs(t); Min_dis[u] = min(Min_dis[u], Min_dis[t]);
	}
}
int main(){
	T=read();
	while(T--){ Init_graph();
		memset(f,0,sizeof(f));
		memset(val,0,sizeof(val));
		n=read(), m=read();
		for(int i=1;i<=n*2;i++) fa[i]=i;
		for(int i=1;i<=m;i++){
			int x=read(),y=read(),z=read(),a=read();
			E[i] = (Node){x,y,a}; 
			add(x,y,z); add(y,x,z);
		}
		dijsktra();
		Kruskal(); 
		for(int i=1;i<=n*2;i++) Min_dis[i] = inf;
		dfs(cnt);
		
		q=read(),k=read(),s=read();
		ans = 0;
		while(q--){
			int x=read(),y=read();
			x = (LL)(x + k* ans -1) % n + 1;
			y = (LL)(y + k* ans ) % (s+1);
			for(int i=20;i>=0;i--)
				if(f[x][i] && val[f[x][i]]>y) x = f[x][i];
			ans = Min_dis[x];
			printf("%lld\n",ans);
		}
	}return 0;
}

标签：cnt,ch,int,Kruskal,tot,归程,NOI2018,dis
来源： https://blog.csdn.net/sslz_fsy/article/details/86623656