1 条题解

  • -1
    @ 2024-1-8 1:25:10

    树链剖分+树上差分

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 5e4+10;
    int n, k;
    vector<int> g[N];
    int fa[N], top[N], son[N], dep[N], sz[N], cnt[N], ans;
    void dfs1(int x, int father){
    	fa[x] = father, dep[x] = dep[father] + 1,sz[x] = 1;
    	for (auto k : g[x]){
    		if(k == father) continue;
    		dfs1(k,x);
    		sz[x] += sz[k];
    		if(sz[son[x]] < sz[k]) son[x] = k;
    	}
    }
    void dfs2(int x, int t){
    	top[x] = t;
    	int s = son[x];
    	if(!s) return;
    	dfs2(s,t);
    	for(auto k : g[x]){
    		if(k == fa[x] || k == s) continue;
    		dfs2(k,k);
    	}
    }
    int lca(int u, int v){
    	while(top[u] != top[v]){
    		if(dep[top[u]] < dep[top[v]]) swap(u,v);
    		u = fa[top[u]];
    	}
    	return dep[u] < dep[v] ? u : v;
    }
    
    int dfs(int x, int father){
    	int tot = cnt[x];
    	for(auto k : g[x]){
    		if(k == father) continue;
    		tot += dfs(k, x);
    	}
    	ans = max(ans, tot);
    	return tot;
    }
    int main(){
    	cin >> n >> k;
    	for (int i =1 ; i < n; i ++){
    		int a, b;
    		cin >> a >> b;
    		g[a].push_back(b);
    		g[b].push_back(a);
    	}
    	dfs1(1,0);
    	dfs2(1,1);
    	while(k --){
    		int a, b;
    		cin >> a >> b;
    		int lc = lca(a,b);
    		cnt[a] ++ , cnt[b] ++, cnt[lc] --, cnt[fa[lc]] --;
    	}
    	dfs(1,0);
    	cout << ans;
            return 0;
    }
    
    • 1

    信息

    ID
    2967
    时间
    1000ms
    内存
    256MiB
    难度
    10
    标签
    (无)
    递交数
    6
    已通过
    5
    上传者