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 ", 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 ", 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 ",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"); }