Codeforces 741 D - Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths

D - Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths

思路：

树上启发式合并

从根节点出发到每个位置的每个字符的奇偶性记为每个位置的状态，每次统计一下每个状态的最大深度

为了保证链经过当前节点u，我们先计算每个子树的答案，再更新子树状态对深度的贡献。

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
//#define mp make_pair
#define pb push_back
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdi pair<double, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

inline int read() {
    int a = 1, b = 0;
    char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘) {
        if(ch == ‘-‘) a = -1;
        ch = getchar();
    }
    while(‘0‘ <= ch && ch <= ‘9‘) {
        b = b*10 + ch-‘0‘;
        ch = getchar();
    }
    return a*b;
}
const int N = 5e5 + 5, M = 5e6 + 5;
const int INF = 1e8;
vector<pii> g[N];
int n, p, dp[N], sz[N], son[N], deep[N], st[N], mx[M];
char c[2];
void get_son(int u, int o) {
    sz[u] = 1;
    deep[u] = deep[o] + 1;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        int w = g[u][i].se;
        st[v] = st[u] ^ (1<<w);
        get_son(v, u);
        if(sz[v] > sz[son[u]]) son[u] = v;
        sz[u] += sz[v];
    }
}
void CAL(int p, int u) {
    if(mx[st[u]] >= 0) dp[p] = max(dp[p], mx[st[u]]+deep[u]-2*deep[p]);
    for (int i = 0; i < 22; ++i) {
        if(mx[st[u]^(1<<i)] >= 0) dp[p] = max(dp[p], mx[st[u]^(1<<i)]+ deep[u]-2*deep[p]);
    }
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        CAL(p, v);
    }
}
void ADD(int u) {
    mx[st[u]] = max(mx[st[u]], deep[u]);
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        ADD(v);
    }
}
void DELETE(int u) {
    if(mx[st[u]] >= 0) mx[st[u]] = -INF;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        DELETE(v);
    }
}
void dfs(int u) {
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        if(v != son[u]) {
            dfs(v);
            DELETE(v);
        }
    }

    if(son[u]) dfs(son[u]);

    if(mx[st[u]] >= 0) dp[u] = mx[st[u]] - deep[u];
    for (int i = 0; i < 22; ++i) {
        if(mx[st[u]^(1<<i)] >= 0) dp[u] = max(dp[u], mx[st[u]^(1<<i)] - deep[u]);
    }

    mx[st[u]] = max(mx[st[u]], deep[u]);
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        if(v != son[u]) {
            CAL(u, v);
            ADD(v);
        }
    }
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        dp[u] = max(dp[u], dp[v]);
    }
}
int main() {
    n = read();
    for (int i = 2; i <= n; ++i) {
        p = read();
        scanf("%s", c);
        g[p].pb({i, c[0]-‘a‘});
    }
    get_son(1, 0);
    for (int i = 0; i < M; ++i) mx[i] = -INF;
    dfs(1);
    for (int i = 1; i <= n; ++i) printf("%d%c", dp[i], " \n"[i==n]);
    return 0;
}

原文地址：https://www.cnblogs.com/widsom/p/10773406.html