并查集练习

1. luogu P4185 MooTube

大意: 给定树, 定义两点距离为两点树链上的最小边权, m个询问(k,v), 求到v距离>=k的点的个数.

 

离线后用并查集不断添边即可, 在线的话可以用kruskal重构树

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define pb push_back
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;

const int N = 1e5+10;
int n, q, s[N], cnt[N], ans[N];
struct _ {int u,v,w;} e[N];
struct __ {int k,v,id;} qry[N];
int Find(int x) {return s[x]?s[x]=Find(s[x]):x;}
void add(int x, int y) {
	x = Find(x), y = Find(y);
	if (x!=y) s[x]=y,cnt[y]+=cnt[x];
}
int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,n-1) scanf("%d%d%d", &e[i].u,&e[i].v,&e[i].w);
	sort(e+1,e+n,[](_ a,_ b){return a.w>b.w;});
	REP(i,1,q) scanf("%d%d", &qry[i].k, &qry[i].v),qry[i].id=i;
	sort(qry+1,qry+1+q,[](__ a,__ b){return a.k>b.k;});
	REP(i,1,n) cnt[i] = 1;
	int now = 1;
	REP(i,1,q) {
		while (now<=n-1&&e[now].w>=qry[i].k) { 
			add(e[now].u,e[now].v),++now;
		}
		ans[qry[i].id] = cnt[Find(qry[i].v)];
	}
	REP(i,1,q) printf("%d\n", ans[i]-1);
}

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define pb push_back
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;

const int N = 1e5+10;
int n, q, tot, s[N<<1], val[N<<1];
vector<int> g[N<<1];
int sz[N<<1], fa[N<<1][20];
struct _ {int u,v,w;} e[N];
int Find(int x) {return s[x]?s[x]=Find(s[x]):x;}
void dfs(int x, int f) {
	sz[x] = 1;
	fa[x][0] = f;
	REP(i,1,19) fa[x][i]=fa[fa[x][i-1]][i-1];
	for (int y:g[x]) dfs(y,x),sz[x]+=sz[y];
}
int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,n-1) scanf("%d%d%d", &e[i].u,&e[i].v,&e[i].w);
	sort(e+1,e+n,[](_ a,_ b){return a.w>b.w;});
	int tot = n;
	REP(i,1,n-1) {
		int u=Find(e[i].u),v=Find(e[i].v);
		s[u] = s[v] = ++tot, val[tot]=e[i].w;
		g[tot].pb(u), g[tot].pb(v);
	}
	dfs(tot,0);
	REP(i,1,q) {
		int k, v;
		scanf("%d%d", &k, &v);
		PER(i,0,19) if (val[fa[v][i]]>=k) v=fa[v][i];
		printf("%d\n", sz[v]>>1);
	}
}

 

2. hdu 3938 Portal

大意: 无向图, 求有多少条路径, 满足路径上边权的最大值<=L.

可以离线后用并查集算出每条边的贡献.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const int N = 1e6+10;
int n, m, q, s[N], sz[N];
struct _ {int u,v,w;} e[N];
pii a[N];
ll ans[N];
int Find(int x) {return s[x]?s[x]=Find(s[x]):x;}
ll add(int x, int y) {
	x=Find(x), y=Find(y);
	if (x==y) return 0;
	ll t = (ll)sz[x]*sz[y];
	s[x] = y, sz[y] += sz[x];
	return t;
}

void work() {
	REP(i,1,m) scanf("%d%d%d", &e[i].u,&e[i].v,&e[i].w);
	sort(e+1,e+1+m,[](_ a,_ b){return a.w<b.w;});
	REP(i,1,q) scanf("%d",&a[i].x),a[i].y=i;
	sort(a+1,a+1+q);
	REP(i,1,n) sz[i] = 1, s[i] = 0;
	int now = 1;
	REP(i,1,q) {
		ans[a[i].y] = ans[a[i-1].y];
		while (now<=m&&e[now].w<=a[i].x) { 
			ans[a[i].y] += add(e[now].u,e[now].v);
			++now;
		}
	}
	REP(i,1,q) printf("%lld\n",ans[i]);
}

int main() {
	for (; ~scanf("%d%d%d", &n, &m, &q); ) work();
}

 

3. CF 891C Envy

大意: 给定无向图, 每次询问给出一个边集S, 求判断S中的所有边是否能在某个最小生成树上.

对于一条边(u,v,w), 若(u,v,w)能在一个MST上, 那么必须满足所有<w的边权不能连通点u和v, 所有边集离线后跑一次kruskal即可. 

#include <iostream>
#include <algorithm>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;

const int N = 1e6+10;
int n, m, q, tot, clk;
int ans[N], fa1[N], fa2[N], c[N];
struct _ {
	int u,v,w,id;
	bool operator < (const _ &rhs) const {
		return w<rhs.w||w==rhs.w&&id<rhs.id;
	}
} e[N], s[N];
int Find1(int x) {return fa1[x]?fa1[x]=Find1(fa1[x]):x;}
void add(int x, int y) {
	x=Find1(x),y=Find1(y);
	if (x!=y) fa1[x]=y;
}
int Find2(int x) {
	if (c[x]!=clk) fa2[x]=fa1[x],c[x]=clk;
	return fa2[x]?fa2[x]=Find2(fa2[x]):x;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
	scanf("%d", &q);
	REP(i,1,q) {
		int k, t;
		scanf("%d", &k);
		REP(j,1,k) {
			scanf("%d", &t);
			++tot, s[tot] = e[t], s[tot].id = i;
		}
	}
	sort(e+1,e+1+m), sort(s+1,s+1+tot);
	int now = 1;
	for (int i=1; i<=tot; ) {
		while (e[now].w<s[i].w) add(e[now].u,e[now].v),++now;
		++clk;
		for (int t=s[i].w; s[i].w==t; ++i) {
			if (s[i].id!=s[i-1].id) ++clk;
			int u=Find2(s[i].u),v=Find2(s[i].v);
			if (u==v) ans[s[i].id]=1;
			else fa2[u]=v;
		}
	}
	REP(i,1,q) puts(ans[i]?"NO":"YES");
}