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.
离线后并查集合并即可, 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); } }
原文:https://www.cnblogs.com/uid001/p/10698311.html