int n, m, sz[N], dep[N], mi[N]; int L[N], R[N], fa[N], son[N], top[N]; vector<int> g[N], gg[N]; int s[N], cnt, vis[N]; void dfs(int x, int d, int f) { L[x]=++*L,sz[x]=1,fa[x]=f,dep[x]=d; for (int y:g[x]) if (y!=f) { dfs(y,d+1,x),sz[x]+=sz[y]; if (sz[y]>sz[son[x]]) son[x]=y; } R[x]=*L; } void dfs(int x, int tf) { top[x]=tf; if (son[x]) dfs(son[x],tf); for (int y:g[x]) if (!top[y]) dfs(y,y); } int lca(int x, int y) { while (top[x]!=top[y]) { if (dep[top[x]]<dep[top[y]]) swap(x,y); x=fa[top[x]]; } return dep[x]<dep[y]?x:y; } bool cmp(int x, int y) { return L[x]<L[y]; } void solve(vector<int> a) { sort(a.begin(),a.end(),cmp); int sz = a.size(); REP(i,1,sz-1) a.pb(lca(a[i],a[i-1])); sort(a.begin(),a.end(),cmp); a.erase(unique(a.begin(),a.end()),a.end()); s[cnt=1]=a[0],sz=a.size(); REP(i,1,sz-1) { while (cnt>=1) { if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) { gg[s[cnt]].pb(a[i]); break; } --cnt; } s[++cnt]=a[i]; } DP(s[1]); for (int x:a) gg[x].clear(); }
例1 luogu P2495 [SDOI2011]消耗战
大意: 给定有根树树, 边有边权, 根为$1$, $m$个询问, 每次给出$k$个点, 询问使根节点与$k$个点不连通要删除的边权和的最小值.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #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 hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 111; #endif int n, m, sz[N], dep[N], mi[N]; int L[N], R[N], fa[N], son[N], top[N]; struct _ {int to,w;}; vector<_> g[N]; vector<int> gg[N]; int s[N], cnt, vis[N]; void dfs(int x, int d, int f, int m) { L[x]=++*L,sz[x]=1,fa[x]=f,dep[x]=d,mi[x]=m; for (_ e:g[x]) if (e.to!=f) { int y=e.to; dfs(y,d+1,x,min(m,e.w)),sz[x]+=sz[y]; if (sz[y]>sz[son[x]]) son[x]=y; } R[x]=*L; } void dfs(int x, int tf) { top[x]=tf; if (son[x]) dfs(son[x],tf); for (_ e:g[x]) if (!top[e.to]) dfs(e.to,e.to); } int lca(int x, int y) { while (top[x]!=top[y]) { if (dep[top[x]]<dep[top[y]]) swap(x,y); x=fa[top[x]]; } return dep[x]<dep[y]?x:y; } bool cmp(int x, int y) { return L[x]<L[y]; } ll DP(int x) { if (vis[x]) return 1e17; ll ans = 0; for (int y:gg[x]) ans+=min(DP(y),(ll)mi[y]); return ans; } void solve(vector<int> a) { sort(a.begin(),a.end(),cmp); int sz = a.size(); REP(i,1,sz-1) a.pb(lca(a[i],a[i-1])); a.pb(1); sort(a.begin(),a.end(),cmp); a.erase(unique(a.begin(),a.end()),a.end()); s[cnt=1]=a[0],sz=a.size(); REP(i,1,sz-1) { while (cnt>=1) { if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) { gg[s[cnt]].pb(a[i]); break; } --cnt; } s[++cnt]=a[i]; } printf("%lld\n", DP(1)); for (int x:a) gg[x].clear(); } int main() { scanf("%d", &n); REP(i,1,n-1) { int u, v, w; scanf("%d%d%d", &u, &v, &w); g[u].pb({v,w}),g[v].pb({u,w}); } dfs(1,0,0,INF),dfs(1,1); scanf("%d", &m); REP(i,1,m) { int k, t; scanf("%d", &k); vector<int> v; REP(i,1,k) scanf("%d",&t),v.pb(t),vis[t]=1; solve(v); for (int x:v) vis[x]=0; } }