树链剖分

推荐博客:https://www.cnblogs.com/ivanovcraft/p/9019090.html

前置知识:

dfs序,线段树

主要应用:树上有关问题的维护,将书上问题转化为序列问题从而用线段树进行统计维护

大概过程:

1,dfs1计算Size[x]数组(表示x这个树的大小),d数组(表示结点深度),son[x]数组(表示x的子结点中最大的结点),far[x]数组(表示x结点的父亲结点)

void dfs1(int u,int f,int dep)///dfs1指在处理d数组,son数组,far数组,Size数组
{
    d[u] = dep; far[u] = f;
    Size[u] = 1; son[u] = -1;
    for(int i = head[u]; i; i = Next[i]){
        int v = ver[i];
        if(v == f) continue;
        dfs1(v,u,dep+1);
        Size[u] += Size[v];
        if(son[u] == -1 || Size[son[u]] < Size[v])
           son[u] = v;
    }
}

 2,dfs2计算dfs序列,旨在处理好重链

void dfs2(int u,int T)///旨在处理重链,和dfs序列
{
    dfn[++ cnt] = u;id[u] = cnt;
    top[u] = T;
    if(son[u] == -1) return ;
    dfs2(son[u],T);
    for(int i = head[u]; i; i = Next[i]){
        int v = ver[i];
        if(v != son[u] && v != far[u]){
            dfs2(v,v);
        }
    }
}

 通过这两个dfs,我们就可以将树上的每个结点转化为序列上的结点了。

例题:https://loj.ac/problem/10138

不知道出于什么想法,一开始一直觉得不用建树,然后wa wawawawa

#include "stdio.h"
#include "string.h"
#include "algorithm"
using namespace std;

const int N = 60010, M = 70010;
const int INF = -3001010;

int n, q;
int head[N], ver[N], Next[N], tot;  ///树的结构存储
int val[N];                         ///存储每个结点的信息
int d[N], son[N], far[N], Size[N];  ///结点的深度,重儿子,祖先
int a[N], maxx[N * 4], sum[N * 4];  ///线段树上的结点值,maxx,sum值
int dfn[N], top[N], id[N];  ///存储dfs序,top是条链的祖先,id是每个结点在dfn中序列的下标位置
int cnt;                    ///表示的是dfs序列的最后一个位置

void add(int x, int y) {  ///添加树边
    ver[++tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
}

void Build_Tree(int id, int l, int r) {
    if (l == r) {
        sum[id] = maxx[id] = val[dfn[l]];
        return;
    }
    int mid = (l + r) >> 1;
    Build_Tree(id * 2, l, mid);
    Build_Tree(id * 2 + 1, mid + 1, r);
    sum[id] = sum[id * 2] + sum[id * 2 + 1];
    maxx[id] = max(maxx[id * 2], maxx[id * 2 + 1]);
    return;
}
void Update(int id, int l, int r, int loc, int x)  ///将loc上的值进行更新
{
    if (l == r) {
        maxx[id] = x;
        sum[id] = x;
        return;
    }
    int mid = (l + r) >> 1;
    if (loc <= mid)
        Update(id * 2, l, mid, loc, x);
    else
        Update(id * 2 + 1, mid + 1, r, loc, x);
    maxx[id] = max(maxx[id * 2], maxx[id * 2 + 1]);
    sum[id] = 0;
    if (sum[id * 2] != INF)
        sum[id] += sum[id * 2];
    if (sum[id * 2 + 1] != INF)
        sum[id] += sum[id * 2 + 1];
}

int Query_sum(int id, int L, int R, int l, int r)  ///查询[l,r]区间和
{
    if (L > r || R < l)
        return 0;
    if (l <= L && r >= R) {
        return sum[id];
    }
    int mid = (L + R) >> 1;
    int ans = Query_sum(id * 2, L, mid, l, r) + Query_sum(id * 2 + 1, mid + 1, R, l, r);
    return ans;
}

int Query_maxx(int id, int L, int R, int l, int r)  ///查询区间[l,r]之间的最大值
{
    if (l <= L && r >= R)
        return maxx[id];
    int mid = (L + R) >> 1;
    int ans = -3 * 10010;
    if (l <= mid)
        ans = max(ans, Query_maxx(id * 2, L, mid, l, r));
    if (r > mid)
        ans = max(ans, Query_maxx(id * 2 + 1, mid + 1, R, l, r));
    return ans;
}

void dfs1(int u, int f, int dep)  /// dfs1指在处理d数组,son数组,far数组,Size数组
{
    d[u] = dep;
    far[u] = f;
    Size[u] = 1;
    son[u] = -1;
    for (int i = head[u]; i; i = Next[i]) {
        int v = ver[i];
        if (v == f)
            continue;
        dfs1(v, u, dep + 1);
        Size[u] += Size[v];
        if (son[u] == -1 || Size[son[u]] < Size[v])
            son[u] = v;
    }
}

void dfs2(int u, int T)  ///旨在处理重链,和dfs序列
{
    dfn[++cnt] = u;
    id[u] = cnt;
    top[u] = T;
    if (son[u] == -1)
        return;
    dfs2(son[u], T);
    for (int i = head[u]; i; i = Next[i]) {
        int v = ver[i];
        if (v != son[u] && v != far[u]) {
            dfs2(v, v);
        }
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y);
        add(y, x);
    }
    cnt = 0;
    dfs1(1, 0, 1);
    dfs2(1, 1);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &val[i]);
    }
    Build_Tree(1, 1, n);
    scanf("%d", &q);
    while (q--) {
        char str[15];
        int u, v;
        scanf("%s%d%d", str, &u, &v);
        if (!strcmp(str, "QMAX"))  ///求最大值
        {
            int fu = top[u], fv = top[v];
            int ans = -300100;
            while (fu != fv) {
                if (d[fu] >= d[fv]) {
                    ans = max(ans, Query_maxx(1, 1, n, id[fu], id[u]));
                    u = far[fu];
                    fu = top[u];
                } else {
                    ans = max(ans, Query_maxx(1, 1, n, id[fv], id[v]));
                    v = far[fv];
                    fv = top[v];
                }
            }
            if (id[u] <= id[v])
                ans = max(ans, Query_maxx(1, 1, n, id[u], id[v]));
            else
                ans = max(ans, Query_maxx(1, 1, n, id[v], id[u]));
            printf("%d\n", ans);
        }
        if (!strcmp(str, "QSUM"))  ///求和
        {
            int fu = top[u], fv = top[v];
            int ans = 0;
            while (fu != fv) {
                if (d[fu] >= d[fv]) {
                    ans += Query_sum(1, 1, n, id[fu], id[u]);
                    u = far[fu];
                    fu = top[u];
                } else {
                    ans += Query_sum(1, 1, n, id[fv], id[v]);
                    v = far[fv];
                    fv = top[v];
                }
            }
            if (id[u] < id[v])
                ans += Query_sum(1, 1, cnt, id[u], id[v]);
            else
                ans += Query_sum(1, 1, cnt, id[v], id[u]);
            printf("%d\n", ans);
        }
        if (!strcmp(str, "CHANGE"))  ///更新指定u位置上的值更新为v
        {
            Update(1, 1, cnt, id[u], v);
        }
    }
}
//

 

posted @ 2020-02-23 00:48  风生  阅读(160)  评论(0编辑  收藏  举报