hdu 6203 ping ping ping(LCA+树状数组) hdu 6203 ping ping ping(LCA+树状数组)

hdu 6203 ping ping ping(LCA+树状数组)
hdu 6203 ping ping ping(LCA+树状数组)

题意:给一棵树,有m条路径,问至少删除多少个点使得这些路径都不连通

(1 <= n <= 1e4)
(1 <= m <= 5e4)

思路:

根据路径的LCA深度从大到小排序,每次选择一个没被删除的LCA删除
当某个点删除时,跨越了以这个点为根的子树的路径都会被割断,而排序保证在同一子树内部的路径已经被处理过了,子树信息可以用dfs序来表示,区间操作可以用左右端点打标记,用树状数组维护即可。

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 1e4 + 10;
vector<int> G[N];
int tot,in[N],out[N];
int dep[N],f[N][20];
int n;
void dfs(int u,int fa,int d){
    in[u] = ++tot;
    f[u][0] = fa,dep[u] = d;
    for(int i = 1;i < 20;i++) f[u][i] = f[f[u][i-1]][i-1];
    for(auto v:G[u]) if(v != fa) dfs(v,u,d + 1);
    out[u] = tot;
}
int lca(int u,int v){
    if(dep[u] < dep[v]) swap(u,v);
    int d = dep[u] - dep[v];
    for(int i = 19;i >= 0 && u != v;i--) if(d & (1<<i)) u = f[u][i];
    if(u == v) return u;
    for(int i = 19;i >= 0;i--) if(f[u][i] != f[v][i]) u = f[u][i],v = f[v][i];
    return f[u][0];
}
int tr[N];
int lowbit(int x){return x &(-x);}
void up(int pos,int c){
    for(;pos <= n;pos += lowbit(pos)) tr[pos] += c;
}
int getsum(int pos){
    int ans = 0;
    for(;pos;pos -= lowbit(pos)) ans += tr[pos];
    return ans;
}
struct Q{
    int x,y,xy;
    Q(){};
    bool operator<(const Q&rhs)const{
        return dep[xy] > dep[rhs.xy];
    }
}qr[100010];
int main()
{
    int u,v;
    while(scanf("%d",&n)==1){
        n++;
        for(int i = 1;i <= n;i++) {
                G[i].clear();
                tr[i] = 0;
        }
        for(int i = 1;i < n;i++){
            scanf("%d%d",&u,&v);u++,v++;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        tot = 0;
        dfs(1,0,0);
        int q;
        scanf("%d",&q);
        for(int i = 0;i < q;i++){
            scanf("%d%d",&u,&v);u++,v++;
            qr[i].x = u,qr[i].y = v,qr[i].xy = lca(u,v);
        }
        sort(qr, qr + q);
        int ans = 0;
        for(int i = 0;i < q;i++){
            if(getsum(in[qr[i].x]) || getsum(in[qr[i].y])) continue;
            ans++;up(in[qr[i].xy],1),up(out[qr[i].xy]+1,-1);
        }
        cout<<ans<<endl;
    }