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  • Split The Tree

    Problem E. Split The Tree

    Time Limit: 7000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 409    Accepted Submission(s): 109


    Problem Description
    You are given a tree with n vertices, numbered from 1 to n. ith vertex has a value wi
    We define the weight of a tree as the number of different vertex value in the tree.
    If we delete one edge in the tree, the tree will split into two trees. The score is the sum of these two trees’ weights.
    We want the know the maximal score we can get if we delete the edge optimally
     
    Input
    Input is given from Standard Input in the following format:
    n
    p2 p3 . . . pn
    w1 w2 . . . wn
    Constraints
    2 ≤ n ≤ 100000
    1 ≤ pi < i
    1 ≤ wi ≤ 100000(1 ≤ i ≤ n), and they are integers
    pi means there is a edge between pi and i
     
    Output
    Print one number denotes the maximal score
     
    Sample Input
    3 1 1 1 2 2 3 1 1 1 1 1
     
    Sample Output
    3 2
     
    Source
     
    #include <bits/stdc++.h>
    
    using namespace std;
    const int maxn = 6e5 + 10;
    int n;
    int val[maxn], dfn[maxn], l[maxn], r[maxn], tot, cnt, nx[maxn], to[maxn], c[maxn], vis[maxn];
    vector<int> G[maxn];
    
    struct node {
        int L, R, pos;
    
        bool operator<(const node &cur) const {
            if (L == cur.L)return R < cur.R;
            return L < cur.L;
        }
    
    } o[maxn];
    
    inline void dfs(int cur, int f) {
        l[cur] = ++tot;
        dfn[tot] = val[cur];
        for (register int i = 0; i < G[cur].size(); ++i) {
            int to = G[cur][i];
            if (to == f)continue;
            dfs(to, cur);
        }
        r[cur] = tot;
    }
    
    inline int lowbit(int x) {
        return x & (-x);
    }
    
    inline void update(int pos, int val) {
        while (pos <= n) {
            c[pos] += val;
            pos += lowbit(pos);
        }
    }
    
    inline int query(int pos) {
        int res = 0;
        while (pos) {
            res += c[pos];
            pos -= lowbit(pos);
        }
        return res;
    }
    
    int q[maxn];
    
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("1.txt", "r", stdin);
    #endif
        while (scanf("%d", &n) != EOF) {
            tot = 0;
            memset(q, 0, sizeof(q));
            for (register int i = 1; i <= n; ++i) {
                G[i].clear();
            }
            for (register int i = 2, pi; i <= n; ++i) {
                scanf("%d", &pi);
                G[pi].emplace_back(i);
                G[i].emplace_back(pi);
            }
            int mx = -1;
            for (register int i = 1; i <= n; ++i) {
                scanf("%d", &val[i]);
                mx = max(mx, val[i]);
            }
            dfs(1, -1);
            cnt = 0;
            for (register int i = 0; i <= mx; ++i) {
                to[i] = 2 * n + 1;
                vis[i] = 0;
            }
            for (register int i = 1; i <= n; ++i) {
                dfn[i + n] = dfn[i];
                o[++cnt] = node{l[i], r[i], i};
                o[++cnt] = node{r[i] + 1, l[i] - 1 + n, i};
            }
            sort(o + 1, o + 1 + cnt);
            n *= 2;
    
            for (register int i = n; i >= 1; --i) {
                nx[i] = to[dfn[i]];
                to[dfn[i]] = i;
                c[i] = 0;
            }
            for (register int i = 1; i <= n; ++i) {
                if (!vis[dfn[i]]) {
                    vis[dfn[i]] = 1;
                    update(i, 1);
                }
            }
            int Left = 1, res = 0;
            for (register int i = 1; i <= cnt; ++i) {
                while (Left < o[i].L) {
                    if (nx[Left] <= n)update(nx[Left], 1);
                    ++Left;
                }
                q[o[i].pos] += query(o[i].R) - query(o[i].L - 1);
                res = max(res, q[o[i].pos]);
            }
            printf("%d
    ", res);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/czy-power/p/11433485.html
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