BZOJ 1036: [ZJOI2008]树的统计Count 【树链剖分】

1036: [ZJOI2008]树的统计Count

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 10395  Solved: 4215
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Description

一棵树上有n个节点，编号分别为1到n，每个节点都有一个权值w。我们将以下面的形式来要求你对这棵树完成一些操作： I. CHANGE u t : 把结点u的权值改为t II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值 III. QSUM u v: 询问从点u到点v的路径上的节点的权值和 注意：从点u到点v的路径上的节点包括u和v本身

Input

输入的第一行为一个整数n，表示节点的个数。接下来n – 1行，每行2个整数a和b，表示节点a和节点b之间有一条边相连。接下来n行，每行一个整数，第i行的整数wi表示节点i的权值。接下来1行，为一个整数q，表示操作的总数。接下来q行，每行一个操作，以“CHANGE u t”或者“QMAX u v”或者“QSUM u v”的形式给出。 对于100％的数据，保证1<=n<=30000，0<=q<=200000；中途操作中保证每个节点的权值w在-30000到30000之间。

Output

对于每个“QMAX”或者“QSUM”的操作，每行输出一个整数表示要求输出的结果。

Sample Input

4
1 2
2 3
4 1
4 2 1 3
12
QMAX 3 4
QMAX 3 3
QMAX 3 2
QMAX 2 3
QSUM 3 4
QSUM 2 1
CHANGE 1 5
QMAX 3 4
CHANGE 3 6
QMAX 3 4
QMAX 2 4
QSUM 3 4

Sample Output

4
1
2
2
10
6
5
6
5
16

HINT

很裸的树链剖分了，不多说了。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (30000+10)#define MAXM (300000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while((a)--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)#pragma comment(linker, "/STACK:102400000,102400000")#define fi first#define se secondusing namespace std;struct Tree{    int l, r, sum, Max;};Tree tree[MAXN<<2];void PushUp(int o){    tree[o].sum = tree[ll].sum + tree[rr].sum;    tree[o].Max = max(tree[ll].Max, tree[rr].Max);}void Build(int o, int l, int r){    tree[o].l = l; tree[o].r = r;    tree[o].sum = tree[o].Max = 0;    if(l == r)        return ;    int mid = (l + r) >> 1;    Build(lson); Build(rson);}void Update(int o, int pos, int v){    if(tree[o].l == tree[o].r)    {        tree[o].sum = tree[o].Max = v;        return ;    }    int mid = (tree[o].l + tree[o].r) >> 1;    if(pos <= mid) Update(ll, pos, v);    else Update(rr, pos, v);    PushUp(o);}int Query(int o, int L, int R, int op){    if(tree[o].l == L && tree[o].r == R)        return op == 1 ? tree[o].sum : tree[o].Max;    int mid = (tree[o].l + tree[o].r) >> 1;    if(R <= mid) return Query(ll, L, R, op);    else if(L > mid) return Query(rr, L, R, op);    else return op == 1 ? Query(ll, L, mid, op) + Query(rr, mid+1, R, op) : max(Query(ll, L, mid, op), Query(rr, mid+1, R, op));}struct Edge{    int from, to, next;};Edge edge[MAXN<<1];int head[MAXN], edgenum;void init(){    edgenum = 0; CLR(head, -1);}void addEdge(int u, int v){    Edge E = {u, v, head[u]};    edge[edgenum] = E;    head[u] = edgenum++;}int son[MAXN], num[MAXN];int top[MAXN], pos[MAXN], id;int dep[MAXN], pre[MAXN];void DFS1(int u, int fa, int d){    dep[u] = d; pre[u] = fa; num[u] = 1; son[u] = -1;    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        if(v == fa) continue;        DFS1(v, u, d+1);        num[u] += num[v];        if(son[u] == -1 || num[son[u]] < num[v])            son[u] = v;    }}void DFS2(int u, int T){    top[u] = T; pos[u] = ++id;    if(son[u] == -1) return ;    DFS2(son[u], T);    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        if(v == pre[u] || v == son[u]) continue;        DFS2(v, v);    }}int Get(int u, int v, int op){    int f1 = top[u], f2 = top[v];    int ans;    if(op == 1) ans = 0;    else ans = -INF;    while(f1 != f2)    {        if(dep[f1] < dep[f2])        {            swap(u, v);            swap(f1, f2);        }        if(op == 2) ans = max(ans, Query(1, pos[f1], pos[u], op));        else ans += Query(1, pos[f1], pos[u], op);        u = pre[f1], f1 = top[u];    }    if(dep[u] > dep[v]) swap(u, v);    if(op == 2) ans = max(ans, Query(1, pos[u], pos[v], op));    else ans += Query(1, pos[u], pos[v], op);    return ans;}int a[MAXN];int main(){    int n;    while(Ri(n) != EOF)    {        init();        for(int i = 1; i <= n-1; i++)        {            int s, e;            Ri(s), Ri(e);            addEdge(s, e);            addEdge(e, s);        }        DFS1(1, -1, 1); id = 0; DFS2(1, 1); Build(1, 1, id);        for(int i = 1; i <= n; i++) Ri(a[i]);        for(int i = 1; i <= n; i++) Update(1, pos[i], a[i]);        int m; Ri(m);        W(m)        {            char op[5];            Rs(op); int x, y;            Ri(x); Ri(y);            if(op[1] == 'M')                Pi(Get(x, y, 2));            else if(op[1] == 'S')                Pi(Get(x, y, 1));            else                Update(1, pos[x], y);        }    }    return 0;}