Codeforces 1111E DP + 树状数组 + LCA + dfs序

痛定思痛。 2021-12-12 01:57 286阅读 0赞

题意：给你一颗树，有q次询问，每次询问给你若干个点，这些点可以最多分出m组，每组要满足两个条件：1：每组至少一个点，2：组内的点不能是组内其它点的祖先，问这样的分组能有多少个？

思路：[https://blog.csdn.net/BUAA\_Alchemist/article/details/86765501][https_blog.csdn.net_BUAA_Alchemist_article_details_86765501]

代码：

#include <bits/stdc++.h>
    #define LL long long
    #define lowbit(x) (x & (-x))
    using namespace std;
    const LL mod = 1000000007;
    const int maxn = 100010;
    vector<int> G[maxn];
    vector<int> a;
    int dfn[maxn], sz[maxn], tot, t;
    LL dp[maxn][310];
    int n;
    void add(int x, int y) {
    	G[x].push_back(y);
    	G[y].push_back(x);
    }
    void dfs(int x, int fa) {
    	dfn[x] = ++tot;
    	sz[x] = 1;
    	for (auto y : G[x]) {
    		if(y == fa) continue;
    		dfs(y, x);
    		sz[x] += sz[y];
    	}
    }
    queue<int> q;
    int dep[maxn], f[maxn][20];
    void bfs() {
    	q.push(1);
    	dep[1] = 1;
    	while(q.size()) {
    		int x = q.front();
    		q.pop();
    		for (auto y : G[x]) {
    			if(dep[y]) continue;
    			dep[y] = dep[x] + 1;
    			//dis[y] = dis[x] + 1;
    			f[y][0] = x;
    			for (int j = 1; j <= t; j++)
    				f[y][j] = f[f[y][j - 1]][j - 1];
    			q.push(y);
    		}
    	}
    }
    
    int lca(int x, int y) {
    	if(dep[x] > dep[y]) swap(x, y);
    	for (int i = t; i >= 0; i--)
    		if(dep[f[y][i]] >= dep[x]) y = f[y][i];
    	if(x == y) return y;
    	for (int i = t; i >= 0; i--)
    		if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    	return f[x][0];
    }
    struct BIT {
    	int c[maxn];
    	int ask(int x) {
    		int ans = 0;
    		for(; x; x -= lowbit(x)) ans += c[x];
    		return ans;
    	}
    	
    	void add(int x, int y) {
    		for(; x <= n; x += lowbit(x)) c[x] += y;
    	}
    	
    };
    BIT tr;
    int h[maxn], vis[maxn];
    int main() {
    	int u, v, T;
    	scanf("%d%d", &n, &T);
    	t = (int)(log(n) / log(2)) + 1;
    	for (int i = 1; i < n; i++) {
    		scanf("%d%d", &u, &v);
    		add(u, v);
    	}
    	dfs(1, -1);
    	bfs();
    	int k, m, r, x;
    	LL ans = 0;
    	while(T--) {
    		scanf("%d%d%d",&k, &m, &r);
    		ans = 0;
    		for (int i = 1; i <= k; i++) {
    			scanf("%d", &x);
    			vis[x] = 1;
    			a.push_back(x);
    			tr.add(dfn[x], 1);
    			tr.add(dfn[x] + sz[x], -1);
    		}
    		for (int i = 0; i < k; i++) {
    			int LCA = lca(a[i], r);
    			h[i + 1] = tr.ask(dfn[a[i]]) + tr.ask(dfn[r]) - 2 * tr.ask(dfn[LCA]) + vis[LCA] - 1;
    		}
    		sort(h + 1, h + 1 + k);
    		dp[0][0] = 1;
    		for (int i = 1; i <= k; i++)
    			for (int j = 0; j <= min(i, m); j++) {
    				if(j > 0)
    					dp[i][j] = (LL)((LL)dp[i - 1][j - 1] + ((LL)dp[i - 1][j] * max(0, j - h[i])) % mod) % mod;
    			}
    		for (int i = 1; i <= m; i++)
    			ans = (ans + dp[k][i]) % mod;
    		printf("%lld\n", ans);
    		for (int i = 0; i < k; i++) {
    			tr.add(dfn[a[i]], -1);
    			tr.add(dfn[a[i]] + sz[a[i]], 1);
    			vis[a[i]] = 0;
    		}
    		a.clear();
    	}
    }

转载于:https://www.cnblogs.com/pkgunboat/p/10918148.html

[https_blog.csdn.net_BUAA_Alchemist_article_details_86765501]: https://blog.csdn.net/BUAA_Alchemist/article/details/86765501