HDU - 5390 tree 线段树套字典树 （看题解）

HDU - 5390

看到的第一感觉就是树链剖分 + 线段树套字典树， 感觉复杂度不太对。

其实这种路径其实很特殊， 一个点改变只会影响它儿子到根的路径， 并且这种求最优值问题可以叠加。

所以我们修改的时候对对应dfs序打标记， 询问的时候在线段树上从上往下对每个对应区间求个最优值。

这样还会被卡MLE。。 需要分层优化一下。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair
#define PLI pair
#define PII pair
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int LOG = 30;

int n, m, v[N], tmp[N];
int qus[N][3];
int ans[N];
int Trietot;
int ch[N * 40][2], sz[N * 40];
int mx = 0;
int newNode() {
    Trietot++;
    assert(Trietot < N * 40);
    chkmax(mx, Trietot);
    ch[Trietot][0] = 0;
    ch[Trietot][1] = 0;
    sz[Trietot] = 0;
    return Trietot;
}

struct Trie {
    int Rt;
    void init() {
        Rt = newNode();
    }
    void ins(int x) {
        int u = Rt;
        for(int i = LOG - 1; i >= 0; i--) {
            sz[u]++;
            if(!ch[u][x >> i & 1]) {
                ch[u][x >> i & 1] = newNode();
            }
            u = ch[u][x >> i & 1];
        }
        sz[u]++;
    }
    void del(int x) {
        int u = Rt;
        for(int i = LOG - 1; i >= 0; i--) {
            sz[u]--;
            u = ch[u][x >> i & 1];
        }
        sz[u]--;
    }
    int query(int x) {
        int u = Rt, ret = 0;
        if(!sz[u]) return 0;
        for(int i = LOG - 1; i >= 0; i--) {
            if(sz[ch[u][(x >> i & 1) ^ 1]]) {
                ret += 1 << i;
                u = ch[u][(x >> i & 1) ^ 1];
            }
            else {
                u = ch[u][x >> i & 1];
            }
        }
        return ret;
    }
} T[N << 2];


#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
void build(int tar, int cur, int l, int r, int rt) {
    if(cur == tar) {
        T[rt].init();
        return;
    }
    if(l == r) return;
    int mid = l + r >> 1;
    build(tar, cur + 1, lson);
    build(tar, cur + 1, rson);
}
void update(int tar, int cur, int L, int R, int val, int l, int r, int rt) {
    if(L > R) return;
    if(L <= l && r <= R) {
        if(cur == tar) {
            if(val > 0) T[rt].ins(val);
            else T[rt].del(-val);
        }
        return;
    }
    if(l == r) return;
    int mid = l + r >> 1;
    if(L <= mid) update(tar, cur + 1, L, R, val, lson);
    if(R > mid) update(tar, cur + 1, L, R, val, rson);
}

int query(int tar, int cur, int p, int val, int l, int r, int rt) {
    if(cur == tar) {
        return T[rt].query(val);
    }
    if(l == r) return 0;
    int mid = l + r >> 1;
    if(p <= mid) return query(tar, cur + 1, p, val, lson);
    else return query(tar, cur + 1, p, val, rson);
}

void initTrie(int depth) {
    Trietot = 0;
    build(depth, 0, 1, n, 1);
}

vector<int> G[N];
int in[N], ot[N], idx;

void dfs(int u) {
    in[u] = ++idx;
    for(auto &v : G[u]) {
        dfs(v);
    }
    ot[u] = idx;
}

void init() {
    idx = 0;
    for(int i = 1; i <= n; i++) {
        G[i].clear();
    }
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        init();
        for(int i = 2; i <= n; i++) {
            int par; scanf("%d", &par);
            G[par].push_back(i);
        }
        for(int i = 1; i <= n; i++) {
            scanf("%d", &v[i]);
            tmp[i] = v[i];
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &qus[i][0], &qus[i][1]);
            if(qus[i][0] == 0) scanf("%d", &qus[i][2]);
            ans[i] = 0;
        }
        dfs(1);
        for(int depth = 0; depth <= 17; depth++) {
            initTrie(depth);

            for(int i = 1; i <= n; i++) {
                v[i] = tmp[i];
                update(depth, 0, in[i], ot[i], v[i], 1, n, 1);
            }
            for(int i = 1; i <= m; i++) {
                if(!qus[i][0]) {
                    update(depth, 0, in[qus[i][1]], ot[qus[i][1]], -v[qus[i][1]], 1, n, 1);
                    v[qus[i][1]] = qus[i][2];
                    update(depth, 0, in[qus[i][1]], ot[qus[i][1]], v[qus[i][1]], 1, n, 1);
                }
                else {
                    chkmax(ans[i], query(depth, 0, in[qus[i][1]], v[qus[i][1]], 1, n, 1));
                }
            }
        }

        for(int i = 1; i <= m; i++) {
            if(qus[i][0]) {
                printf("%d\n", ans[i]);
            }
        }
    }
    return 0;
}

/*
*/

 

转载于:https://www.cnblogs.com/CJLHY/p/11183306.html