hdu Dylans loves tree [LCA] (树链剖分)

本文深入探讨了C++模板元编程的概念、原理及其应用,通过具体实例展示了如何利用模板元编程解决实际问题,旨在提升读者在复杂场景下的编程能力。

Dylans loves tree

 

view code#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#define REP(i, n) for(int i = 0; i < (n); ++i)
#define FOR(i, a, b) for(int i = (a); i <= (b); ++i)
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
using namespace std;
typedef vector<int > VI;
typedef pair<int, int > PII;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 100010;
int _, cas=1, n, Q, m, val[N], seg[N << 2];

struct Edge {
    int u, v, next;
    Edge() {}
    Edge(int u, int v, int next):u(u), v(v), next(next) {}
}e[N << 1];
int head[N], ecnt;

void add(int u, int v) {
    e[ecnt] = Edge(u, v, head[u]);
    head[u] = ecnt++;
}

int dep[N], fa[N], sz[N], top[N], id[N], fid[N], gid, son[N];
void dfs(int u, int f, int d) {
    fa[u] = f, sz[u] = 1, dep[u] = d, son[u] = 0;
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].v;
        if(v == f) continue;
        dfs(v, u, d + 1);
        sz[u] += sz[v];
        if(sz[ son[u] ] < sz[v]) son[u] = v;
    }
}

void getpos(int u, int f) {
    id[u] = ++gid;
    fid[gid] = u;
    top[u] = f;
    if(!son[u]) return ;
    getpos(son[u], f);
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].v;
        if(v == son[u] || v == fa[u]) continue;
        getpos(v, v);
    }
}

void pushUp(int rt) {
    seg[rt] = seg[rt << 1] ^ seg[rt << 1 | 1];
}

void build(int l, int r, int rt) {
    if(l == r) {
        seg[rt] = val[ fid[l] ];
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushUp(rt);
}

void update(int p, int c, int l, int r, int rt) {
    if(l == r) {
        seg[rt] = c;
        return ;
    }
    int m = (l + r) >> 1;
    if(p <= m) update(p, c, lson);
    else update(p, c, rson);
    pushUp(rt);
}

int query(int L, int R, int l, int r, int rt) {
    if(L <= l && R >= r) return seg[rt];
    int m = (l + r) >> 1, ans = 0;
    if(L <= m) ans ^= query(L, R, lson);
    if(R > m) ans ^= query(L, R, rson);
    return ans;
}

int lca(int u, int v) {
    int fu = top[u], fv = top[v], ans = 0, dis = 0;
    while(fu != fv) {
        if(dep[fu] < dep[fv]) {
            swap(fu, fv), swap(u, v);
        }
        ans ^= query(id[fu], id[u], 1, n, 1);
        dis += id[u] - id[fu] + 1;
        u = fa[fu];
        fu = top[u];
    }
    if(dep[u] < dep[v]) swap(u, v);
    ans ^= query(id[v], id[u], 1, n, 1);
    dis += id[u] - id[v] + 1;
    return (dis & 1) ? ans : -1;
}

void solve() {
    scanf("%d%d", &n, &Q);

    ecnt = 0;
    memset(head, -1, sizeof(head[0]) * sizeof(n +5));
    int u, v;
    REP(i, n - 1) {
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }
    FOR(i, 1, n) {
        scanf("%d", &val[i]);
    }

    sz[0] = 0;
    gid = 0;
    dfs(1, 1, 0);
    getpos(1, 1);
    build(1, n, 1);

    int flag;
    while(Q--) {
        scanf("%d%d%d", &flag, &u, &v);
        if(flag) {
            printf("%d\n", lca(u, v));
        } else {
            update(id[u], v, 1, n, 1);
        }
    }
}

int main()
{
    cin >> _;
    while(_--) solve();
    return 0;
}

 

转载于:https://www.cnblogs.com/zyx1314/p/4600109.html

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