HYSBZ1036-树的统计Count

1036: [ZJOI2008]树的统计Count

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 18557  Solved: 7561
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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 I
II. 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

解题思路：树链剖分

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m, x, y;int s[100009], nt[100009], e[100009], v[100009], cnt,ss,tt;int ct[100009], mx[100009], fa[100009], dep[100009];int top[100009], g[100009],G[100009],ma[100009<<2];LL sum[100009 << 2];char ch[10];void dfs(int x, int f){dep[x] = dep[f] + 1;fa[x] = f; ct[x] = 1; mx[x] = 0;for (int i = s[x]; ~i; i = nt[i]){if (e[i] == f) continue;dfs(e[i], x);ct[x] += ct[e[i]];if (ct[e[i]] > ct[mx[x]]) mx[x] = e[i];}}void Dfs(int x, int t){top[x] = !t ? x : top[fa[x]];g[x] = ++cnt,G[cnt]=x;if (mx[x]) Dfs(mx[x], 1);for (int i = s[x]; ~i; i = nt[i]){if (e[i] == fa[x]) continue;if (e[i] == mx[x]) continue;Dfs(e[i], 0);}}void build(int k, int l, int r){if (l == r) { sum[k] = ma[k] = v[G[l]]; return; }int mid = (l + r) >> 1;build(k << 1, l, mid);build(k << 1 | 1, mid + 1, r);ma[k] = max(ma[k << 1], ma[k << 1 | 1]);sum[k] = sum[k << 1] + sum[k << 1 | 1];}void update(int k, int l, int r, int p,int val){if (l == r) { sum[k] = ma[k] = val; return; }int mid = (l + r) >> 1;if (p <= mid) update(k << 1, l, mid, p, val);else update(k << 1 | 1, mid + 1, r, p, val);ma[k] = max(ma[k << 1], ma[k << 1 | 1]);sum[k] = sum[k << 1] + sum[k << 1 | 1];}int queryma(int k, int l, int r, int ll, int rr){if (l >= ll&&r <= rr) return ma[k];int mid = (l + r) >> 1, ma = -INF;if (mid >= ll) ma = max(ma, queryma(k << 1, l, mid, ll, rr));if (rr > mid) ma = max(ma, queryma(k << 1 | 1, mid + 1, r, ll, rr));return ma;}int getma(int x, int y){int ma = -INF;while (top[x] != top[y]){if (dep[top[x]] < dep[top[y]]) swap(x, y);ma=max(ma,queryma(1,1,n,g[top[x]],g[x]));x = fa[top[x]];}if (dep[x] > dep[y]) swap(x, y);return max(ma, queryma(1, 1, n, g[x], g[y]));}LL querysum(int k, int l, int r, int ll, int rr){if (l >= ll&&r <= rr) return sum[k];int mid = (l + r) >> 1;LL sum = 0;if (mid >= ll) sum+=querysum(k << 1, l, mid, ll, rr);if (rr > mid) sum+=querysum(k << 1 | 1, mid + 1, r, ll, rr);return sum;}LL getsum(int x, int y){LL sum = 0;while (top[x] != top[y]){if (dep[top[x]] < dep[top[y]]) swap(x, y);sum+=querysum(1, 1, n, g[top[x]], g[x]); x = fa[top[x]];}if (dep[x] > dep[y]) swap(x, y);return sum+querysum(1, 1, n, g[x], g[y]);}int main(){while (~scanf("%d", &n)){memset(s,- 1, sizeof s);dep[0] = ct[0] = cnt = 0;for (int i = 1; i < n; i++){scanf("%d%d", &x, &y);nt[cnt] = s[x], s[x] = cnt,e[cnt++] = y;nt[cnt] = s[y], s[y] = cnt,e[cnt++] = x;}for (int i = 1; i <= n; i++) scanf("%d", &v[i]);dfs(1, 0);Dfs(1, cnt = 0);build(1, 1, n);scanf("%d", &m);while (m--){scanf("%s%d%d", ch, &x, &y);if (ch[0] == 'C') update(1, 1, n, g[x], y);else if (ch[1] == 'M') printf("%d\n",getma(x, y));else printf("%lld\n",getsum(x, y));}}return 0;}