NOI模拟题1 Problem A: sub

题面

Sample Input

5 7
2 -1 -3 1 1
1 2
1 3
3 4
3 5
2
1 3 0
2
1 2 1
2
1 1 -3
2

Sample Output

2
4
5
2

HINT

Solution

首先考虑序列上的这个问题: 给定一个序列, 有两种操作

• 修改一个位置上的值
• 询问某个区间中的连续段的最大权值和

做法是: 用一个线段树维护这个序列, 每个节点所对应的区间记录以下4个信息:

• sum表示整个区间的权值和
• leftMax表示以包含从最左边开始的区间的连续段最大权值和
• rightMax与上面相反
• max表示整个区间中连续段最大权值和

则我们可以通过上述信息合并两个区间的信息.

考虑如何转化到树上解决: 我们首先进行树链剖分. 我们用sum[u]记录\(u\)这个点的权值加上以\(u\)的每个字节点为根的连通块的最大权值和. 我们用类似于序列上的方法维护\(sum\)即可. 修改一个点的权值时, 往上跳修改即可.

实现的时候把线段树上的信息合并重载为运算符会比较简洁.

#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
namespace Zeonfai
{
	inline int getInt()
	{
		int a = 0, sgn = 1; char c;
		while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
		while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
		return a * sgn;
	}
}
const int N = (int)1e5;
int n;
multiset<int> ans;
struct data
{
	int leftMax, rightMax, mx, sum;
	inline data() { leftMax = rightMax = mx = sum = 0; }
	inline data friend operator +(const data &a, const data &b)
	{
		data res;
		res.sum = a.sum + b.sum;
		res.leftMax = max(a.leftMax, a.sum + b.leftMax);
		res.rightMax = max(b.rightMax, b.sum + a.rightMax);
		res.mx = max(res.leftMax, res.rightMax);
		res.mx = max(res.mx, a.mx); res.mx = max(res.mx, b.mx);
		res.mx = max(res.mx, a.rightMax + b.leftMax);
		return res;
	}
};
struct segmentTree
{
	data nd[N << 2];
	void modify(int u, int L, int R, int pos, int x)
	{
		if (L == R)
		{
			nd[u].leftMax = nd[u].rightMax = nd[u].mx = max(0, x);
			nd[u].sum = x;
			return;
		}
		if (pos <= L + R >> 1) modify(u << 1, L, L + R >> 1, pos, x);
		else modify(u << 1 | 1, (L + R >> 1) + 1, R, pos, x);
		nd[u] = nd[u << 1] + nd[u << 1 | 1];
	}
	inline void modify(int pos, int x) { modify(1, 1, n, pos, x); }
 	data query(int u, int curL, int curR, int L, int R)
	{
		if (curL >= L && curR <= R) return nd[u];
		int mid = curL + curR >> 1;
		data res;
		if (L <= mid) res = res + query(u << 1, curL, mid, L, R);
		if (R > mid) res = res + query(u << 1 | 1, mid + 1, curR, L, R);
		return res;
	}
	inline data query(int L, int R) { return query(1, 1, n, L, R); }
}seg;
struct tree
{
	struct node
	{
		int w;
		vector<int> edg; int pre;
		int sz, hvy;
		int tp, id, lst;
		int sum;
		inline node() { edg.clear(); }
	}nd[N + 1];
	inline void addEdge(int u, int v) { nd[u].edg.push_back(v); nd[v].edg.push_back(u); }
	void getSize(int u, int pre)
	{
		nd[u].pre = pre; nd[u].hvy = -1; nd[u].sz = 1;
		for (auto v : nd[u].edg) if (v != pre)
		{
			getSize(v, u); nd[u].sz += nd[v].sz;
			if (nd[u].hvy == -1 || nd[v].sz > nd[nd[u].hvy].sz) nd[u].hvy = v;
		}
	}
	int clk;
	int work(int u, int tp)
	{
		nd[u].tp = tp; nd[u].id = ++ clk;
		if (~ nd[u].hvy) work(nd[u].hvy, tp);
		else
		{
			for (int v = u; v != nd[u].tp; v = nd[v].pre) nd[v].lst = nd[u].id;
			nd[nd[u].tp].lst = nd[u].id;
		}
		nd[u].sum = nd[u].w;
		for (auto v : nd[u].edg) if (v != nd[u].pre && v != nd[u].hvy) nd[u].sum += work(v, v);
		seg.modify(nd[u].id, nd[u].sum);
		if (u == tp)
		{
			data res = seg.query(nd[u].id, nd[u].lst);
			ans.insert(res.mx);
			return res.leftMax;
		}
		return 0;
	}
	inline void decomposition() { getSize(1, -1); clk = 0; work(1, 1); }
	inline void modify(int u, int x)
	{
		int tmp = nd[u].w; nd[u].w = x;
		for (; ~ u; u = nd[nd[u].tp].pre)
		{
			nd[u].sum = nd[u].sum - tmp + x;
			data res = seg.query(nd[nd[u].tp].id, nd[nd[u].tp].lst);
			tmp = res.leftMax;
			ans.erase(ans.find(res.mx));
			seg.modify(nd[u].id, nd[u].sum);
			res = seg.query(nd[nd[u].tp].id, nd[nd[u].tp].lst);
			x = res.leftMax;
			ans.insert(res.mx);
		}
	}
}T;
int main()
{
#ifndef ONLINE_JUDGE
	freopen("sub.in", "r", stdin);
	freopen("sub.out", "w", stdout);
#endif
	using namespace Zeonfai;
	n = getInt(); int m = getInt();
	for (int i = 1; i <= n; ++ i) T.nd[i].w = getInt();
	for (int i = 1, u, v; i < n; ++ i) u = getInt(), v = getInt(), T.addEdge(u, v);
	ans.clear();
	T.decomposition();
	for (int i = 0; i < m; ++ i)
	{
		int opt = getInt();
		if (opt == 1)
		{
			int u = getInt(), x = getInt();
			T.modify(u, x);
		}
		else
		{
			multiset<int>::iterator p = ans.end();
			printf("%d\n", *(-- p));
		}
	}
}