HDU3669 Aragorn's Story 树链剖分 点权

HDU3669 Aragorn's Story 树链剖分 点权

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3966

题意:

n个点的,m条边,每个点都 有点权

修改 从u->v上所有点的点权

查询单点点权

题解:

树链剖分裸题

树链剖分就是将树分割为多条边,然后利用数据结构来维护这些链的一个技巧

  • 重儿子:父亲节点的所有儿子中子树结点数目最多( sizesiz**e 最大)的结点;
  • 轻儿子:父亲节点中除了重儿子以外的儿子;
  • 重边:父亲结点和重儿子连成的边;
  • 轻边:父亲节点和轻儿子连成的边;
  • 重链:由多条重边连接而成的路径;
  • 轻链:由多条轻边连接而成的路径;

1.求出子树大小和每个点的重儿子,处理sz数组,son数组,fa数组和dep数组

2.连接重链,记录dfs序,每个链的顶端节点,处理出rank数组,top数组,id数组

3.维护链上信息

代码:

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;

int n, m, Q;
int a[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn], id[maxn], son[maxn], Rank[maxn];
int sum[maxn << 2], lazy[maxn << 2];

struct EDGE {
    int u, v, nt;
} edge[maxn << 1];
int head[maxn], summ, cnt;

void add_edge(int u, int v) {
    edge[++summ].u = u; edge[summ].v = v; edge[summ].nt = head[u]; head[u] = summ;
}

void dfs1(int u) {
    sz[u] = 1; son[u] = 0;
    for (int i = head[u]; ~i; i = edge[i].nt) {
        int v = edge[i].v;
        if (v != fa[u]) {
            fa[v] = u;
            dep[v] = dep[u] + 1;
            dfs1(v);
            sz[u] += sz[v];
            if (sz[v] > sz[son[u]]) son[u] = v;
        }
    }
}

void dfs2(int u, int tp, int x) {
    top[u] = tp; id[u] = ++cnt; Rank[cnt] = u;
    if (son[u]) dfs2(son[u], tp, 1);
    for (int i = head[u]; ~i; i = edge[i].nt) {
        int v = edge[i].v;
        if (v == son[u] || v == fa[u]) continue;
        dfs2(v, v, 2);
    }
}

void init() {
    memset(head, -1, sizeof(head));
    summ = 1; cnt = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        add_edge(u, v); add_edge(v, u);
    }
    dep[1] = 1; fa[1] = 0;
    dfs1(1);
    dfs2(1, 1, 1);
}

void pushup(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void pushdown(int rt, int mid) {
    if (lazy[rt]) {
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        sum[rt << 1] += lazy[rt] * (mid - mid / 2);
        sum[rt << 1 | 1] += lazy[rt] * (mid / 2);
        lazy[rt] = 0;
    }
}

void build(int l, int r, int rt) {
    lazy[rt] = 0;
    if (l == r) {
        sum[rt] = a[Rank[l]];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int L, int R, int val, int l, int r, int rt) {
    if (L <= l && r <= R) {
        sum[rt] += val * (r - l + 1);
        lazy[rt] += val;
        return;
    }
    pushdown(rt, r - l + 1);
    int mid = (l + r) >> 1;
    if (L <= mid) update(L, R, val, lson);
    if (mid < R) update(L, R, val, rson);
    pushup(rt);
}

int query(int pos, int l, int r, int rt) {
    if (l == r) {
        return sum[rt];
    }
    pushdown(rt, r - l + 1);
    int mid = (l + r) >> 1;
    if (pos <= mid) return query(pos, lson);
    if (mid < pos) return query(pos, rson);
}

void change(int x, int y, int val) {
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) std::swap(x, y);
        update(id[top[x]], id[x], val, 1, n, 1);
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) std::swap(x, y);
    update(id[x], id[y], val, 1, n, 1);
}

int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    while (~scanf("%d %d %d", &n, &m, &Q)) {
        init();
        build(1, n, 1);
        while (Q--) {
            char s[2];
            int x, y, z;
            scanf("%s", s);
            if (s[0] == 'I') {
                scanf("%d %d %d", &x, &y, &z);
                change(x, y, z);
            }
            if (s[0] == 'D') {
                scanf("%d %d %d", &x, &y, &z);
                change(x, y, -z);
            }
            if (s[0] == 'Q') {
                scanf("%d", &x);
                printf("%d
", query(id[x], 1, n, 1));
            }
        }
    }
}
每一个不曾刷题的日子 都是对生命的辜负 从弱小到强大,需要一段时间的沉淀,就是现在了 ~buerdepepeqi
原文地址:https://www.cnblogs.com/buerdepepeqi/p/11203995.html