BZOJ 1036 树的统计Count(树链剖分(点权)入门题)

题意：给你一个n个节点的一棵树，每个节点有一个权值，现在q次操作,操作有三种:

Change x y : 将x节点的权值改成y

QMAX x y : 问你x到y路径上的最大值

QSUM x y: 问你x到y路径上的权值和

思路：第一个树剖题，树剖裸题，剖完后直接线段树查询即可。

num[]数组，用来保存以x为根的子树节点个数
top[]数组，用来保存当前节点的所在链的顶端节点
son[]数组，用来保存重儿子
deep[]数组，用来保存当前节点的深度
fa[]数组，用来保存当前节点的父亲
tree[]数组，用来保存树中每个节点剖分后的新编号
pre[]数组，用来保存线段树当前点对应树节点的编号

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;typedef long long ll;const int maxn = 1e5+5;const int INF = 0x3f3f3f3f;int deep[maxn], fa[maxn], top[maxn], tree[maxn], pre[maxn], son[maxn], num[maxn];int n, tot, treeSum[maxn*4], treeMax[maxn*4], a[maxn];vector<int> g[maxn];void init(){    tot = 0;    memset(son, 0, sizeof(son));    memset(num, 0, sizeof(num));    for(int i = 0; i <= n; i++) g[i].clear();}void dfs1(int u, int pre,  int d){    deep[u] = d;    fa[u] = pre;    num[u] = 1;    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(v == pre) continue;        dfs1(v, u, d+1);        num[u] += num[v];        if(!son[u] || num[v] > num[son[u]])            son[u] = v;    }}void dfs2(int u, int tp){    top[u] = tp;    tree[u] = ++tot;    pre[tree[u]] = u;    if(!son[u]) return ;    dfs2(son[u], tp);    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(v != son[u] && v != fa[u])            dfs2(v, v);    }}void push_up(int root){    treeMax[root] = max(treeMax[root*2], treeMax[root*2+1]);    treeSum[root] = treeSum[root*2]+treeSum[root*2+1];}void build(int root, int l, int r){    if(l == r)    {        treeMax[root] = a[pre[l]];        treeSum[root] = a[pre[l]];        return ;    }    int mid = (l+r)/2;    build(root*2, l, mid);    build(root*2+1, mid+1, r);    push_up(root);}void update(int root, int l, int r, int pos, int val){    if(l == r)    {        treeMax[root] = val;        treeSum[root] = val;        return ;    }    int mid = (l+r)/2;    if(pos <= mid) update(root*2, l, mid, pos, val);    else update(root*2+1, mid+1, r, pos, val);    push_up(root);}int querySum(int root, int l, int r, int i, int j){    if(i <= l && j >= r)        return treeSum[root];    int mid = (l+r)/2, sum = 0;    if(i <= mid) sum += querySum(root*2, l, mid, i, j);    if(j > mid) sum += querySum(root*2+1, mid+1, r, i, j);    return sum;}int queryMax(int root, int l, int r, int i, int j){    if(i <= l && j >= r)        return treeMax[root];    int mid = (l+r)/2;    if(j <= mid) return queryMax(root*2, l, mid, i, j);    else if(i > mid) return queryMax(root*2+1, mid+1, r, i, j);    else return max(queryMax(root*2, l, mid, i, j), queryMax(root*2+1, mid+1, r, i, j));}int askSum(int x, int y){    int f1= top[x], f2 = top[y], ans = 0;    while(f1 != f2)    {        if(deep[f1] < deep[f2]) swap(f1, f2), swap(x, y);        ans += querySum(1, 1, n, tree[f1], tree[x]);        x = fa[f1]; f1 = top[x];    }    ans += (deep[x]>deep[y]) ? querySum(1, 1, n, tree[y], tree[x]) : querySum(1, 1, n, tree[x], tree[y]);    return ans;}int askMax(int x, int y){    int f1 = top[x], f2 = top[y], ans = -INF;    while(f1 != f2)    {        if(deep[f1] < deep[f2]) swap(f1, f2), swap(x, y);        ans = max(ans, queryMax(1, 1, n, tree[f1], tree[x]));        x = fa[f1]; f1 = top[x];    }    ans = max(ans, (deep[x]>deep[y]) ? queryMax(1, 1, n, tree[y], tree[x]) : queryMax(1, 1, n, tree[x], tree[y]));    return ans;}int main(void){    while(cin >> n)    {        init();        for(int i = 1; i < n; i++)        {            int u, v;            scanf("%d%d", &u, &v);            g[u].push_back(v);            g[v].push_back(u);        }        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]);        dfs1(1, 0, 1); dfs2(1, 1);        build(1, 1, n);        int q;        char cmd[15];        scanf("%d", &q);        while(q--)        {            int x, y;            scanf(" %s%d%d", cmd, &x, &y);            if(cmd[0] == 'C') update(1, 1, n, tree[x], y);            else if(cmd[1] == 'M') printf("%d\n", askMax(x, y));            else printf("%d\n", askSum(x, y));        }    }    return 0;}